RD Sharma 2022 Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

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Page No 168:

Question 1:

Which of the following are quadratic equations?

(i) 16x² − 3 = (2x + 5) (5x − 3)

(ii) $\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0$

(iii) $x^{2} + \frac{1}{x^{2}} = 5$

(iv) $x - \frac{3}{x} = x^{2}$

(v) $2 x^{2} - \sqrt{3 x} + 9 = 0$

(vi) $x^{2} - 2 x - \sqrt{x} - 5 = 0$

â€‹(vii) (x + 2)³ = x³ − 4

(viii) $x + \frac{1}{x} = 1$

(ix) $x + \frac{1}{x} = x^{2}, x \neq 0$

(x) ${(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$ â€‹

Answer:

We are given the following algebraic expressions and are asked to find out which one is quadratic.

(i) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and.

Hence, the above equation is a quadratic equation.

(ii) Here it has been given that,

Now, solving the above equation further we get,

Now, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(iii) Here it has been given that,

Now, solving the above equation further we get,

$\frac{x^{4} + 1}{x^{2}} = 5 \Rightarrow x^{4} + 1 = 5 x^{2} \Rightarrow x^{4} - 5 x^{2} + 1 = 0$

Now, the above equation clearly does not represent a quadratic equation of the form , because $x^{4} - 5 x^{2} + 1$ is a polynomial of degree 4.

Hence, the above equation is not a quadratic equation.

(iv) Here it has been given that,

Now, solving the above equation further we get,

Now, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial of degree 3.

Hence, the above equation is not a quadratic equation.

(v) Here it has been given that,

Now, the above equation clearly does not represent a quadratic equation of the form, because contains a term , where is not an integer.

Hence, the above equation is not a quadratic equation.

(vi) Here it has been given that,

Now, as we can see the above equation clearly does not represent a quadratic equation of the form, because contains an extra term, where is not an integer.

Hence, the above equation is not a quadratic equation.

(vii) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(viii) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(ix) Here it has been given that,

, $x \neq 0$

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial having a degree of 3 which is never present in a quadratic polynomial.

Hence, the above equation is not a quadratic equation.

(x) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial having a degree of 4 which is never present in a quadratic polynomial.

Hence, the above equation is not a quadratic equation.

Page No 169:

Question 2:

In each of the following, determine whether the given values are solutions of the given equation or not:

(i) $x^{2} - 3 \sqrt{3} x + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}$

(ii) $x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{4}{3}$

(iii) $2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3$

(iv) $x^{2} - \sqrt{2} x - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2}$

(v) $a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}$

(vi) $\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 16}, x = 3$ â€‹

Answer:

We are given the following quadratic equations and we are asked to find whether the given values are solutions or not

(i) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that and are the solutions of the given quadratic equation.

(ii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence, is not a solution of the quadratic equation.

Therefore, from the above results we find out that both and are not the solutions of the given quadratic equation.

(iii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the given quadratic equation

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that both and are solutions of the quadratic equation.

(iv) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Therefore, from the above results we find out that is a solution but is not a solution of the given quadratic equation.

(v) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that is not a solution and is a solution of the given quadratic equation.

(vi) We have been given that,

$\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 16}, x = 3$

Now, if $x = 3$ is a solution of the equation, then it should satisfy the equation.

So, substituting $x = 3$ in the equation, we get
$\begin{array}{rcl} \sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} & = & \sqrt{3^{2} - 4 (3) + 3} + \sqrt{3^{2} - 9} \\ = & \sqrt{9 - 12 + 3} + \sqrt{9 - 9} \\ = & 0 \end{array}$
$\begin{array}{rcl} \sqrt{4 x^{2} - 14 x + 16} & = & \sqrt{4 {(3)}^{2} - 14 (3) + 16} \\ = & \sqrt{36 - 42 + 16} \\ = & \sqrt{10} \neq 0 \end{array}$

$\Rightarrow \sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} \neq \sqrt{4 x^{2} - 14 x + 16}$ at $x = 3$

Hence, $x = 3$ is not a solution of the quadratic equation.

Page No 169:

Question 3:

In each of the following, find the value of k for which the given value is a solution of the given equation:

(i) $7 x^{2} + k x - 3 = 0, x = \frac{2}{3}$
(ii) $x^{2} - x (a + b) + k = 0, x = a$
(iii) $k x^{2} + \sqrt{2} x - 4 = 0, x = \sqrt{2}$
(iv) $x^{2} + 3 a x + k = 0, x = - a$

Answer:

In each of the following cases find k.

(i) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of .

(ii) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of .

(iii) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of.

(iv) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence the value of.

Page No 169:

Question 4:

If x = 2/3 and x = −3 are the roots of the equation ax² + 7x + b = 0, find the values of a and b.

Answer:

We have been given that,

We have to find a and b

Now, if is a root of the equation, then it should satisfy the equation completely. Therefore we substitute in the above equation. We get,

…… (1)

Also, if is a root of the equation, then it should satisfy the equation completely. Therefore we substitute in the above equation. We get,

…… (2)

Now, we multiply equation (2) by 9 and then subtract equation (1) from it. So we have,

Now, put this value of ‘a’ in equation (2) in order to get the value of ‘b’. So,

Therefore, we have and .

Page No 171:

Question 1:

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.

Answer:

Since it is given in the question that the numbers we have to find are consecutive positive integer numbers, therefore the difference between the two numbers should be equal to 1.
For e.g. 7 and 8 or 26 and 27 are consecutive numbers.

Let us assume the first number to be ‘x’. So our next consecutive number should be ‘x + 1’. Now the question also says that the product of these two numbers is 306.

Therefore,

Hence, this is our required quadratic equation.

Page No 171:

Question 2:

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John and x marbles.

Answer:

It is given that John had ‘x’ marbles.

We are also given that both John and Javanti had 45 marbles together.
So, Javanti should have ’45 − x’ marbles with her.

Now, it is given that both of them lose 5 marbles each.

So in the new situation, John will have ‘x − 5’marbles and Javanti will have ’45 − x − 5’ marbles.

Also it is given that the product of the number of marbles both of them now is 128.

Therefore,

Hence, this is the required quadratic equation.

Page No 171:

Question 3:

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.

Answer:

Now, since we have to find out base, let us assume the base to be ‘x’ cm.
Therefore the height of the triangle becomes ‘x −7’.

It is also given that the hypotenuse is 13 cm.
By Pythagoras Theorem,

Hence, this is our required quadratic equation.

Page No 171:

Question 4:

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, from the quadratic equation of find x.

Answer:

Now we know that ‘x’ denotes the total number of toys produced in that day.

But, the cost of production of a single toy is 55 minus the number of toys produced that day i.e. ‘x’.

So, the total production cost would be the product of the cost of a single toy and the total number of toys i.e. product of ‘55 − x’ and ‘x’. Now, it is given here that total production cost of that day was Rs.750.

Therefore,

Hence, this is the required quadratic equation.

Page No 172:

Question 5:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Banglore. If the average speed of the express train is 11 km/hr more than that of the passenger train, from the quadratic equation to find the average speed of express train.

Answer:

Now let us assume that the speed of the express train be ‘x’ km/hr. Therefore according to the question speed of the passenger train will be ‘x −11’ km/hr. Now we know that the total distance travelled by both the trains was 132 km.

We also know that

So the time taken by express train would be hr and the time taken by the passenger train would behr. Now, we also know that the express train took 1 hr less than the passenger train to travel the whole distance.

Therefore, we have

Therefore, this is the required equation.

Page No 172:

Question 6:

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.

Answer:

Let us assume that the speed of the train be ‘x’ km/hr. We are also given that the distance covered during the journey is 360 km.

Now, time taken during the journey = hr

Time taken for the new journey = hr

According to the question,

Hence, this is the required quadratic equation.

Page No 180:

Question 1:

Solve the following quadratic equations by factorization:

25x (x + 1) = −4

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 2:

Solve the following quadratic equations by factorization:

$16 x - \frac{10}{x} = 27$

Answer:

$16 x - \frac{10}{x} = 27 \Rightarrow 16 x^{2} - 10 = 27 x \Rightarrow 16 x^{2} - 27 x - 10 = 0 \Rightarrow 16 x^{2} - 32 x + 5 x - 10 = 0 \Rightarrow 16 x (x - 2) + 5 (x - 2) = 0 \Rightarrow (16 x + 5) (x - 2) = 0 \Rightarrow 16 x + 5 = 0 or x - 2 = 0 \Rightarrow x = - \frac{5}{16} or x = 2$

Hence, the factors are $2$ and $- \frac{5}{16}$ .

Page No 180:

Question 3:

Solve the following quadratic equations by factorization:

6x² + 11x + 3 = 0

Answer:

We have been given

or,

Hence .

Page No 180:

Question 4:

Solve the following quadratic equations by factorization:

$2 x^{2} + a x - a^{2} = 0$

Answer:

$2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (2 x - a) (x + a) = 0 \Rightarrow 2 x - a = 0 or x + a = 0 \Rightarrow x = \frac{a}{2} or x = - a$

Hence, the factors are $\frac{a}{2}$ and $- a$ .

Page No 180:

Question 5:

Solve the following quadratic equations by factorization:

$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5$

Answer:

We have been given

Therefore,

or,

Hence, x = 2 or x = −6.

Page No 180:

Question 6:

Solve the following quadratic equations by factorization:

$\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, x \neq 4, 7$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 7:

Solve the following quadratic equations by factorization:

$\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}; x \neq 0, 2, 3$

Answer:

$\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x} \Rightarrow \frac{(x - 2) + 2 (x - 3)}{(x - 3) (x - 2)} = \frac{8}{x} \Rightarrow \frac{x - 2 + 2 x - 6}{x^{2} - 2 x - 3 x + 6} = \frac{8}{x} \Rightarrow \frac{3 x - 8}{x^{2} - 5 x + 6} = \frac{8}{x} \Rightarrow x (3 x - 8) = 8 (x^{2} - 5 x + 6) \Rightarrow 3 x^{2} - 8 x = 8 x^{2} - 40 x + 48 \Rightarrow 5 x^{2} - 32 x + 48 = 0 \Rightarrow 5 x^{2} - 20 x - 12 x + 48 = 0 \Rightarrow 5 x (x - 4) - 12 (x - 4) = 0 \Rightarrow (5 x - 12) (x - 4) = 0 \Rightarrow 5 x - 12 = 0 or x - 4 = 0 \Rightarrow x = \frac{12}{5} or x = 4$

Hence, the factors are 4 and $\frac{12}{5}$ .

Page No 180:

Question 8:

Solve the following quadratic equations by factorization:

$\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, - 1$

Answer:

$\frac{16}{x} - 1 = \frac{15}{x + 1} \Rightarrow \frac{16 - x}{x} = \frac{15}{x + 1} \Rightarrow (16 - x) (x + 1) = 15 x \Rightarrow 16 x + 16 - x^{2} - x = 15 x \Rightarrow - x^{2} + 16 + 15 x = 15 x \Rightarrow - x^{2} + 16 = 0 \Rightarrow x^{2} - 16 = 0 \Rightarrow (x - 4) (x + 4) = 0 \Rightarrow x - 4 = 0 or x + 4 = 0 \Rightarrow x = 4 or x = - 4$

Hence, the factors are 4 and −4.

Page No 180:

Question 9:

Solve the following quadratic equations by factorization:

$\frac{x + 3}{x + 2} = \frac{3 x - 7}{2 x - 3}$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 10:

Solve the following quadratic equations by factorization:

$\frac{x + 3}{x - 2} - \frac{1 - x}{x} = \frac{17}{4}$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 11:

Solve the following quadratic equations by factorization:

$\frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = \frac{48}{7}, x \neq 3, x \neq - 3$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 12:

Solve the following quadratic equations by factorization:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2}$

Answer:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3} \Rightarrow \frac{4 - 3 x}{x} = \frac{5}{2 x + 3} \Rightarrow (4 - 3 x) (2 x + 3) = 5 x \Rightarrow 8 x + 12 - 6 x^{2} - 9 x = 5 x \Rightarrow - 6 x^{2} - 6 x + 12 = 0 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x - 1) (x + 2) = 0 \Rightarrow x - 1 = 0 or x + 2 = 0 \Rightarrow x = 1 or x = - 2$

Hence, the factors are 1 and −2.

Page No 180:

Question 13:

Solve the following quadratic equations by factorization:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3}$

Answer:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1} \Rightarrow \frac{6 - (x + 1)}{2 (x + 1)} = \frac{2}{3 x - 1} \Rightarrow \frac{6 - x - 1}{2 x + 2} = \frac{2}{3 x - 1} \Rightarrow (5 - x) (3 x - 1) = 2 (2 x + 2) \Rightarrow 15 x - 5 - 3 x^{2} + x = 4 x + 4 \Rightarrow - 3 x^{2} + 16 x - 5 - 4 x - 4 = 0 \Rightarrow - 3 x^{2} + 12 x - 9 = 0 \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow x^{2} - 3 x - x + 3 = 0 \Rightarrow x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 1) (x - 3) = 0 \Rightarrow x - 1 = 0 or x - 3 = 0 \Rightarrow x = 1 or x = 3$

Hence, the factors are 3 and 1.

Page No 180:

Question 14:

Solve the following quadratic equations by factorization:

$\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4 x - 1}; x \neq 1, - 1, \frac{1}{4}$

Answer:

$\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4 x - 1} \Rightarrow \frac{3 (x - 1) + 4 (x + 1)}{(x + 1) (x - 1)} = \frac{29}{4 x - 1} \Rightarrow \frac{3 x - 3 + 4 x + 4}{x^{2} - 1} = \frac{29}{4 x - 1} \Rightarrow \frac{7 x + 1}{x^{2} - 1} = \frac{29}{4 x - 1} \Rightarrow (7 x + 1) (4 x - 1) = 29 (x^{2} - 1) \Rightarrow 28 x^{2} - 7 x + 4 x - 1 = 29 x^{2} - 29 \Rightarrow 29 x^{2} - 28 x^{2} + 3 x - 28 = 0 \Rightarrow x^{2} + 3 x - 28 = 0 \Rightarrow x^{2} + 7 x - 4 x - 28 = 0 \Rightarrow x (x + 7) - 4 (x + 7) = 0 \Rightarrow (x - 4) (x + 7) = 0 \Rightarrow x - 4 = 0 or x + 7 = 0 \Rightarrow x = 4 or x = - 7$

Hence, the factors are 4 and −7.

Page No 180:

Question 15:

Solve the following quadratic equations by factorization:

$\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x}; x \neq 0, - 1, 2$

Answer:

$\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x} \Rightarrow \frac{4 (x - 2) + 3 (x + 1)}{2 (x - 2) (x + 1)} = \frac{23}{5 x} \Rightarrow \frac{4 x - 8 + 3 x + 3}{2 (x^{2} + x - 2 x - 2)} = \frac{23}{5 x} \Rightarrow \frac{7 x - 5}{2 x^{2} - 2 x - 4} = \frac{23}{5 x} \Rightarrow 5 x (7 x - 5) = 23 (2 x^{2} - 2 x - 4) \Rightarrow 35 x^{2} - 25 x = 46 x^{2} - 46 x - 92 \Rightarrow 46 x^{2} - 35 x^{2} - 46 x + 25 x - 92 = 0 \Rightarrow 11 x^{2} - 21 x - 92 = 0 \Rightarrow 11 x^{2} - 44 x + 23 x - 92 = 0 \Rightarrow 11 x (x - 4) + 23 (x - 4) = 0 \Rightarrow (11 x + 23) (x - 4) = 0 \Rightarrow 11 x + 23 = 0 or x - 4 = 0 \Rightarrow x = - \frac{23}{11} or x = 4$

Hence, the factors are 4 and $- \frac{23}{11}$ .

Page No 180:

Question 16:

Solve the following quadratic equations by factorization:

$a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 17:

Solve the following quadratic equations by factorization:

$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0$

Answer:

$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0 \Rightarrow 9 x^{2} - 6 b^{2} x - (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 9 x^{2} + 3 (a^{2} - b^{2}) x - 3 (a^{2} + b^{2}) x - (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 3 x [3 x + (a^{2} - b^{2})] - (a^{2} + b^{2}) [3 x + (a^{2} - b^{2})] = 0 \Rightarrow [3 x - (a^{2} + b^{2})] [3 x + (a^{2} - b^{2})] = 0 \Rightarrow 3 x - (a^{2} + b^{2}) = 0 or 3 x + (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{3} or x = - \frac{a^{2} - b^{2}}{3} \Rightarrow x = \frac{a^{2} + b^{2}}{3} or x = \frac{b^{2} - a^{2}}{3}$

Hence, the factors are $\frac{a^{2} + b^{2}}{3}$ and $\frac{b^{2} - a^{2}}{3}$ .

Page No 180:

Question 18:

Solve the following quadratic equations by factorization:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$

Answer:

We have been given

Therefore,

or,

Hence, or.

Page No 180:

Question 19:

Solve the following quadratic equations by factorization:

$x^{2} + (a + \frac{1}{a}) x + 1 = 0$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 20:

Solve the following quadratic equations by factorization:

$a b x^{2} + (b^{2} - a c) x - b c = 0$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 21:

Solve the following quadratic equations by factorization:
$3 \sqrt{5} x^{2} + 25 x - 10 \sqrt{5} = 0$

Answer:

Consider the equation $3 \sqrt{5} x^{2} + 25 x - 10 \sqrt{5} = 0$
$\Rightarrow 3 \sqrt{5} x^{2} + 30 x - 5 x - 10 \sqrt{5} = 0 \Rightarrow 3 \sqrt{5} x (x + 2 \sqrt{5}) - 5 (x + 2 \sqrt{5}) = 0 \Rightarrow (3 \sqrt{5} x - 5) (x + 2 \sqrt{5}) = 0 \Rightarrow (3 \sqrt{5} x - 5) = 0 or (x + 2 \sqrt{5}) = 0 \Rightarrow x = \frac{\sqrt{5}}{3} or x = - 2 \sqrt{5}$

Page No 180:

Question 22:

Solve the following quadratic equations by factorization:

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

Answer:

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (\sqrt{3} x + \sqrt{2}) (x - \sqrt{6}) = 0 \Rightarrow \sqrt{3} x + \sqrt{2} = 0 or x - \sqrt{6} = 0 \Rightarrow x = - \sqrt{\frac{2}{3}} or x = \sqrt{6}$

Hence, the factors are $\sqrt{6}$ and $- \sqrt{\frac{2}{3}}$ .

Page No 180:

Question 23:

Solve the following quadratic equations by factorization:

$3 x^{2} - 2 \sqrt{6} x + 2 = 0$

Answer:

We have been given

Therefore,

or,

Hence, or .

Page No 180:

Question 24:

Find the roots of the quadratic equation $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ .

Answer:

We write, $7 x = 5 x + 2 x as$ $\sqrt{2} x^{2} \times 5 \sqrt{2} = 10 x^{2} = 5 x \times 2 x$

$∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 \Rightarrow x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \Rightarrow (\sqrt{2} x + 5) (x + \sqrt{2}) = 0$
$\Rightarrow x + \sqrt{2} = 0 or \sqrt{2} x + 5 = 0 \Rightarrow x = - \sqrt{2} or x = - \frac{5}{\sqrt{2}} = - \frac{5 \sqrt{2}}{2}$
Hence, the roots of the given equation are $- \sqrt{2} and$ $- \frac{5 \sqrt{2}}{2}$ .

Page No 180:

Question 25:

Solve the following quadratic equations by factorization:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11; x \neq \frac{3}{5}, - \frac{1}{7}$

Answer:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11 \Rightarrow \frac{3 (7 x + 1)^{2} - 4 {(5 x - 3)}^{2}}{(5 x - 3) (7 x + 1)} = 11 \Rightarrow \frac{3 (49 x^{2} + 1 + 14 x) - 4 (25 x^{2} + 9 - 30 x)}{35 x^{2} + 5 x - 21 x - 3} = 11 \Rightarrow \frac{147 x^{2} + 3 + 42 x - 100 x^{2} - 36 + 120 x}{35 x^{2} - 16 x - 3} = 11 \Rightarrow 47 x^{2} + 162 x - 33 = 11 (35 x^{2} - 16 x - 3) \Rightarrow 47 x^{2} + 162 x - 33 = 385 x^{2} - 176 x - 33 \Rightarrow 385 x^{2} - 47 x^{2} - 176 x - 162 x - 33 + 33 = 0 \Rightarrow 338 x^{2} - 338 x = 0 \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$

Hence, the factors are 0 and 1.

Page No 180:

Question 26:

Solve the following quadratic equations by factorization:

$\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2 x + 3}{x - 2}; x \neq 1, - 2, 2$

Answer:

$\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2 x + 3}{x - 2} \Rightarrow \frac{(x + 1) (x + 2) + (x - 1) (x - 2)}{(x - 1) (x + 2)} = \frac{4 (x - 2) - (2 x + 3)}{x - 2} \Rightarrow \frac{(x^{2} + 2 x + x + 2) + (x^{2} - 2 x - x + 2)}{x^{2} + 2 x - x - 2} = \frac{4 x - 8 - 2 x - 3}{x - 2} \Rightarrow \frac{x^{2} + 3 x + 2 + x^{2} - 3 x + 2}{x^{2} + x - 2} = \frac{2 x - 11}{x - 2}$
$\Rightarrow \frac{2 x^{2} + 4}{x^{2} + x - 2} = \frac{2 x - 11}{x - 2} \Rightarrow (2 x^{2} + 4) (x - 2) = (2 x - 11) (x^{2} + x - 2) \Rightarrow 2 x^{3} - 4 x^{2} + 4 x - 8 = 2 x^{3} + 2 x^{2} - 4 x - 11 x^{2} - 11 x + 22 \Rightarrow 2 x^{3} - 4 x^{2} + 4 x - 8 = 2 x^{3} - 9 x^{2} - 15 x + 22 \Rightarrow 2 x^{3} - 2 x^{3} - 4 x^{2} + 9 x^{2} + 4 x + 15 x - 8 - 22 = 0$
$\Rightarrow 5 x^{2} + 19 x - 30 = 0 \Rightarrow 5 x^{2} + 25 x - 6 x - 30 = 0 \Rightarrow 5 x (x + 5) - 6 (x + 5) = 0 \Rightarrow (5 x - 6) (x + 5) = 0 \Rightarrow 5 x - 6 = 0, x + 5 = 0 \Rightarrow x = \frac{6}{5}, x = - 5$

Page No 180:

Question 27:

Solve the following quadratic equation by factorization:
$\frac{a}{x - b} + \frac{b}{x - a} = 2$

Answer:

$\frac{a}{x - b} + \frac{b}{x - a} = 2$
$\Rightarrow \frac{a x - a^{2} + b x - b^{2}}{(x - a) (x - b)} = 2 \Rightarrow a x - a^{2} + b x - b^{2} = 2 x^{2} - 2 b x - 2 a x + 2 a b \Rightarrow 2 x^{2} - 2 b x - 2 a x + 2 a b - a x + a^{2} - b x + b^{2} = 0 \Rightarrow 2 x^{2} + x [- 2 b - 2 a - a - b] + a^{2} + b^{2} + 2 a b = 0 \Rightarrow 2 x^{2} - 3 x [a + b] + {(a + b)}^{2} = 0$
$\Rightarrow 2 x^{2} - 2 (a + b) x - (a + b) x + {(a + b)}^{2} = 0 \Rightarrow 2 x [x - (a + b)] - (a + b) [x - (a + b)] = 0 \Rightarrow [2 x - (a + b)] [x - (a + b)] = 0$
So, the value of x will be
$x = \frac{a + b}{2}, a + b$

Page No 180:

Question 28:

Solve the following quadratic equations by factorization:

$\frac{a}{x - a} + \frac{b}{x - b} = \frac{2 c}{x - c}$

Answer:

We have been given,

Therefore,

or,

Hence, or .

Page No 180:

Question 29:

Solve the following quadratic equations by factorization:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$

Answer:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 a} = \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{2 a - (2 a + b + 2 x)}{(2 a + b + 2 x) (2 a)} = \frac{2 x + b}{2 b x} \Rightarrow \frac{- b - 2 x}{4 a^{2} + 2 a b + 4 a x} = \frac{2 x + b}{2 b x} \Rightarrow \frac{- 1 (2 x + b)}{4 a^{2} + 2 a b + 4 a x} = \frac{2 x + b}{2 b x} \Rightarrow - 2 b x (2 x + b) = (4 a^{2} + 2 a b + 4 a x) (2 x + b) \Rightarrow (4 a^{2} + 2 a b + 4 a x) (2 x + b) + 2 b x (2 x + b) = 0 \Rightarrow (2 x + b) (4 a^{2} + 2 a b + 4 a x + 2 b x) = 0 \Rightarrow 2 x + b = 0 or 4 a^{2} + 2 a b + (4 a + 2 b) x = 0 \Rightarrow x = - \frac{b}{2} or x = - \frac{4 a^{2} + 2 a b}{4 a + 2 b} \Rightarrow x = - \frac{b}{2} or x = - \frac{a (4 a + 2 b)}{4 a + 2 b} \Rightarrow x = - \frac{b}{2} or x = - a$

Hence, the factors are $- a$ and $- \frac{b}{2}$ .

Page No 180:

Question 30:

Solve the following quadratic equations by factorization:

$(x - 5) (x - 6) = \frac{25}{{(24)}^{2}}$

Answer:

We have been given that,

Therefore,

or,

Hence, or .

Page No 180:

Question 31:

Solve the following quadratic equations by factorization:

$7 x + \frac{3}{x} = 35 \frac{3}{5}$

Answer:

We have been given,

Therefore,

or,

Hence, or .

Page No 185:

Question 1:

Write the discriminant of the following quadratic equations:

(i) (x − 1) (2x − 1) = 0

(ii) $x^{2}$ − 2x + k = 0, k ∈ R

(iii) $\sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0$

(iv) (x + 5)² = 2(5x – 3)

Answer:

We have to find the discriminant of the following quadratic equations

(i) We have been given,

Now, simplify the equation to be represented in the quadratic form, so we have

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(ii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(iii) We have been given,

Now we also know that for an equation , the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(iv) (x + 5)² = 2(5x – 3)

{(x + 5)}^{2} = 2 (5 x - 3) \Rightarrow x^{2} + 5^{2} + 2 (x) (5) = 10 x - 6 \Rightarrow x^{2} + 25 + 10 x = 10 x - 6 \Rightarrow x^{2} + 25 + 10 x - 10 x + 6 = 0 \Rightarrow x^{2} + 31 = 0 Discriminant = b^{2} - 4 a c = 0^{2} - 4 (1) (31) = - 124

Hence, the discriminant is –124.

Page No 185:

Question 2:

â€‹â€‹In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

(i) $3 x^{2} + 2 \sqrt{5} x - 5 = 0$

(ii) $2 x^{2} + 5 \sqrt{3} x + 6 = 0$

(iii) $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$

(iv) $2 x^{2} - 2 \sqrt{2} x + 1 = 0$ â€‹

Answer:

(i) We have been given,

Now we also know that for an equation , the discriminant is given by the following equation: .

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

In order for a quadratic equation to have real roots, . Here, we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are and.

(ii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are and.

(iii) We have been given, .

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are and

(iv) We have been given, .

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is .

Page No 185:

Question 3:

Solve for x:

(i) $\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}; x \neq 2, 4$
(ii) $\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5$
(iii) $x + \frac{1}{x} = 3, x \neq 0$
(iv) $\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1$

(v) â€‹ $\frac{1}{x} + \frac{2}{2 x - 3} = \frac{1}{x - 2}, x \neq 0, \frac{3}{2}, 2$

Answer:

(i) We have been given,

Now we solve the above equation as follows,

Now we also know that for an equation , the discriminant is given by the following equation: .

Now, according to the equation given to us, we have , and .

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of .

(ii) $\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5$

$\Rightarrow \frac{x + 5 - x + 3}{(x - 3) (x + 5)} = \frac{1}{6} \Rightarrow \frac{8}{(x - 3) (x + 5)} = \frac{1}{6}$
$\Rightarrow 48 = x^{2} + 2 x - 15 \Rightarrow x^{2} + 2 x - 15 - 48 = 0 \Rightarrow x^{2} + 2 x - 63 = 0 \Rightarrow x^{2} + 9 x - 7 x - 63 = 0$
$\Rightarrow x (x + 9) - 7 (x + 9) = 0 \Rightarrow (x - 7) (x + 9) = 0 \Rightarrow x = 7, - 9$

(iii) We have been given,

Now, we solve the equation as follows:

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the value of .

(iv) We have been given,

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1$

Now we solve the above equation as follows,

$\frac{16 - x}{x} = \frac{15}{x + 1} \Rightarrow (16 - x) (x + 1) = 15 x \Rightarrow 16 x + 16 - x^{2} - x = 15 x \Rightarrow 15 x + 16 - x^{2} - 15 x = 0 \Rightarrow 16 - x^{2} = 0 \Rightarrow x^{2} - 16 = 0$

Now we also know that for an equation $a x^{2} + b x + c = 0$ , the discriminant is given by the following equation:

$D = b^{2} - 4 a c$

Now, according to the equation given to us, we have, $a = 1$ , $b = 0$ and $c = - 16$ .

Therefore, the discriminant is given as,

$D = (0)^{2} - 4 (1) (- 16) = 64$

Now, the roots of an equation is given by the following equation,

$x = \frac{- b \pm \sqrt{D}}{2 a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{- 0 \pm \sqrt{64}}{2 (1)} = \frac{\pm 8}{2} = \pm 4$

Therefore, the value of $x = \pm 4 .$

(v) $\frac{1}{x} + \frac{2}{2 x - 3} = \frac{1}{x - 2}, x \neq 0, \frac{3}{2}, 2$

$\frac{2 x - 3 + 2 x}{2 x^{2} - 3 x} = \frac{1}{x - 2} \Rightarrow \frac{4 x - 3}{2 x^{2} - 3 x} = \frac{1}{x - 2} \Rightarrow (4 x - 3) (x - 2) = 2 x^{2} - 3 x \Rightarrow 4 x^{2} - 11 x + 6 = 2 x^{2} - 3 x \Rightarrow 2 x^{2} - 8 x + 6 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow (x - 1) (x - 3) = 0 \Rightarrow x = 1 or x = 3$

Page No 191:

Question 1:

Determine the nature of the roots of the following quadratic equations:
â€‹
(i) 2x² − 3x + 5 = 0

(ii) 2x² − 6x + 3 = 0

(iii) $3 x^{2} - 4 \sqrt{3} x + 4 = 0$

(iv) $4 x^{2} + 4 \sqrt{3} x + 3 = 0$

Answer:

(i) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(ii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(v) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iv)

4 x^{2} + 4 \sqrt{3} x + 3 = 0

Discriminant = b^{2} - 4 a c = {(4 \sqrt{3})}^{2} - 4 (4) (3) = 48 - 48 = 0

Since, Discriminant = 0
Therefore, the roots are real and equal.

Hence, the given equation has real and equal roots.

Page No 191:

Question 2:

Find the values of k for which the roots are real and equal in each of the following quadratic equations:

(i) $4 x^{2} - 2 (k + 1) x + (k + 4) = 0$

(ii) $4 x^{2} - 2 (k + 1) x + (k + 1) = 0$

(iii) $x^{2} - 2 (k + 1) x + k^{2} = 0$

(iv) $k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0$

(v) $(k + 1) x^{2} - 2 (k - 1) x + 1 = 0$

(vi) $x^{2} + k (2 x + k - 1) + 2 = 0$ â€‹

Answer:

(i) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(ii) The given equation is $4 x^{2} - 2 (k + 1) x + (k + 1) = 0$ where a = 4, b = −2(k + 1), c = (k + 1)
As we know that

Putting the value of a = 4, b = −2(k + 1), c = (k + 1),
$\begin{array}{rcl} D & = & {\{- 2 (k + 1)\}}^{2} - 4 \times 4 \times (k + 1) \\ = & 4 (k + 1)^{2} - 16 (k + 1) \\ = & (k + 1) \{4 (k + 1) - 16\} \\ = & (k + 1) (4 k - 12) \\ = & 4 (k + 1) (k - 3) \end{array}$

For real and equal roots, D = 0
$\Rightarrow 4 (k + 1) (k - 3) = 0 \Rightarrow k = - 1 or k = 3$

Therefore, the value of k is −1 or 3.

(iii) The given quadric equation is , and roots are real and equal.

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Therefore, the value of

(iv) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Therefore, the value of

(v) The given quadric equation is , and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if D = 0

Now factorizing of the above equation

So, either

Therefore, the value of .

(vi) The given equation is $x^{2} + k (2 x + k - 1) + 2 = 0$ .

$\Rightarrow x^{2} + 2 k x + k (k - 1) + 2 = 0$

So, a = 1, b = 2k, c = k(k − 1) + 2

We know $D = b^{2} - 4 a c$
$\Rightarrow D = {(2 k)}^{2} - 4 \times 1 \times [k (k - 1) + 2] \Rightarrow D = 4 k^{2} - 4 [k^{2} - k + 2] \Rightarrow D = 4 k^{2} - 4 k^{2} + 4 k - 8 \Rightarrow D = 4 k - 8 = 4 (k - 2)$

For equal roots, D = 0
Thus, 4(k − 2) = 0

So, k = 2.

Page No 191:

Question 3:

In each of the following, determine the values of k for which the given quadratic equation has real root:
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
(iii) x²– 4kx + k = 0
â€‹(iv) $k x (x - 2 \sqrt{5}) + 10 = 0$
(v) kx (x – 3) + 9 = 0
(vi) â€‹4x² + kx + 3 = 0

Answer:

(i) The given quadratic equation is 2x² + kx + 3 = 0, and roots are real.

Here, a = 2, b = k, c = 3

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = 2, b = k, c = 3

$D = k^{2} - 4 \times 2 \times 3 = k^{2} - 24 \Rightarrow k^{2} - 24 \geq 0 \Rightarrow k^{2} \geq 24 \Rightarrow k \leq - 2 \sqrt{6} and k \geq 2 \sqrt{6}$

Thus, the value of k is less than $- 2 \sqrt{6}$ and more than $2 \sqrt{6}$ .

(ii) The given quadratic equation is kx (x – 2) + 6 = 0, and roots are real.

kx (x – 2) + 6 = 0
⇒ kx² − 2kx + 6 = 0

Here, a = k, b = −2k, c = 6

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = k, b = −2k, c = 6

$D = {(- 2 k)}^{2} - 4 \times k \times 6 = 4 k^{2} - 24 k \Rightarrow 4 k^{2} - 24 k \geq 0 \Rightarrow k (k - 6) \geq 0 \Rightarrow k \leq 0 and k \geq 6$

Thus, the value of k is less than 0 and more than 6.

(iii) The given quadratic equation is x²– 4kx + k = 0, and roots are real.

Here, a = 1, b = −4k, c = k

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = 1, b = −4k, c = k

$D = {(- 4 k)}^{2} - 4 \times 1 \times k = 16 k^{2} - 4 k \Rightarrow 16 k^{2} - 4 k \geq 0 \Rightarrow 4 k (4 k - 1) \geq 0 \Rightarrow k \leq 0 and k \geq \frac{1}{4}$

Thus, the value of k is less than 0 and more than $\frac{1}{4}$ .

(iv) The given quadratic equation is $k x (x - 2 \sqrt{5}) + 10 = 0$ , and roots are real.

$k x (x - 2 \sqrt{5}) + 10 = 0 \Rightarrow k x^{2} - 2 \sqrt{5} k x + 10 = 0$

Here, a = k, b = − $2 \sqrt{5}$ k, c = 10

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = k, b = − $2 \sqrt{5}$ k, c = 10

$D = {(- 2 \sqrt{5} k)}^{2} - 4 \times k \times 10 = 20 k^{2} - 40 k \Rightarrow 20 k^{2} - 40 k \geq 0 \Rightarrow 20 k (k - 2) \geq 0 \Rightarrow k \leq 0 and k \geq 2$

Thus, the value of k is less than 0 and more than 2.

(v) The given quadratic equation is kx (x – 3) + 9 = 0, and roots are real.

kx (x – 3) + 9 = 0
⇒ kx² − 3kx + 9 = 0

Here, a = k, b = −3k, c = 9

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = k, b = −3k, c = 9

$D = {(- 3 k)}^{2} - 4 \times k \times 9 = 9 k^{2} - 36 k \Rightarrow 9 k^{2} - 36 k \geq 0 \Rightarrow 9 k (k - 4) \geq 0 \Rightarrow k \leq 0 and k \geq 4$

Thus, the value of k is less than 0 and more than 4.

(vi) The given quadratic equation is 4x² + kx + 3 = 0, and roots are real.

Here, a = 4, b = k, c = 3

The given equation will have real roots, if $D = b^{2} - 4 a c$ ≥ 0.

Putting the value of a = 4, b = k, c = 3

$D = {(k)}^{2} - 4 \times 4 \times 3 = k^{2} - 48 \Rightarrow k^{2} - 48 \geq 0 \Rightarrow k^{2} \geq 48 \Rightarrow k \leq - 4 \sqrt{3} and k \geq 4 \sqrt{3}$

Thus, the value of k is less than $- 4 \sqrt{3}$ and more than $4 \sqrt{3}$ .

Page No 192:

Question 4:

Find the values of k for which the given quadratic equation has real and distinct roots:

(a) kx² + 2x + 1 = 0
(b) kx² + 6x + 1 = 0
(c) x² − kx + 9 = 0

Answer:

(i) The given quadric equation is , and roots are real and distinct

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Now according to question, the value of k less than 1

Therefore, the value of

(ii) The given quadric equation is , and roots are real and distinct.

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Now according to question, the value of k less than 9

Therefore, the value of

(iii) The given quadric equation is , and roots are real and distinct

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Therefore, the value of

Page No 192:

Question 5:

For what value of k, (4 − k)x² + (2k + 4) x + (8k + 1) = 0, is a perfect square.

Answer:

The given quadric equation is, and roots are real and equal

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

Page No 192:

Question 6:

(i) Find the values of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has equal roots. Also, find the roots.

(ii) Write all the values of k for which the quadratic equation x² + kx + 16 = 0 has equal roots. Find the roots of the equation so obtained.

Answer:

(i) The given quadric equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ , and roots are real and equal.

Then, find the value of k.

Here, $a = 3 k + 1, b = 2 (k + 1) and c = 1$ .

As we know that $D = b^{2} - 4 a c$

Putting the values of $a = 3 k + 1, b = 2 (k + 1) and c = 1$ .

$D = {[2 (k + 1)]}^{2} - 4 (3 k + 1) (1) = 4 (k^{2} + 2 k + 1) - 12 k - 4 = 4 k^{2} + 8 k + 4 - 12 k - 4 = 4 k^{2} - 4 k$

The given equation will have real and equal roots, if D = 0

Thus, $4 k^{2} - 4 k = 0$
$\Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Therefore, the value of k is 0 or 1.

Now, for k = 0, the equation becomes

$x^{2} + 2 x + 1 = 0 \Rightarrow x^{2} + x + x + 1 = 0 \Rightarrow x (x + 1) + 1 (x + 1) = 0 \Rightarrow (x + 1)^{2} = 0 \Rightarrow x = - 1, - 1$

for k = 1, the equation becomes

$4 x^{2} + 4 x + 1 = 0 \Rightarrow 4 x^{2} + 2 x + 2 x + 1 = 0 \Rightarrow 2 x (2 x + 1) + 1 (2 x + 1) = 0 \Rightarrow (2 x + 1)^{2} = 0 \Rightarrow x = - \frac{1}{2}, - \frac{1}{2}$

Hence, the roots of the equation are $- 1 and - \frac{1}{2}$ .

(ii) x² + kx + 16 = 0

It is given that the quadratic equation has equal roots.
Therefore, Discriminant is equal to zero.

x^{2} + k x + 16 = 0 Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(k)}^{2} - 4 (1) (16) = 0 \Rightarrow k^{2} - 64 = 0 \Rightarrow k^{2} = 64 \Rightarrow k = \pm 8

Hence, the values of k is ±8.

Now,
For k = 8,
The equation becomes
x² + 8x + 16 = 0
⇒ (x + 4)² = 0
⇒ x = −4

For k = −8,
The equation becomes
x² − 8x + 16 = 0
⇒ (x − 4)² = 0
⇒ x = 4

Hence, the roots of the equation so obtained are 4 and −4.

Page No 192:

Question 7:

Find the values of p for which the quadratic equation $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ has equal roots. Also, find these roots.

Answer:

The given quadric equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ , and roots are real and equal.

Then, find the value of p.

Here, $a = 2 p + 1, b = - 7 p - 2 and c = 7 p - 3$ .

As we know that $D = b^{2} - 4 a c$

Putting the values of $a = 2 p + 1, b = - 7 p - 2 and c = 7 p - 3$ .

$D = {[- (7 p + 2)]}^{2} - 4 (2 p + 1) (7 p - 3) = (49 p^{2} + 28 p + 4) - 4 (14 p^{2} - 6 p + 7 p - 3) = 49 p^{2} + 28 p + 4 - 56 p^{2} - 4 p + 12 = - 7 p^{2} + 24 p + 16$

The given equation will have real and equal roots, if D = 0

Thus, $- 7 p^{2} + 24 p + 16 = 0$
$\Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow 7 p^{2} - 28 p + 4 p - 16 = 0 \Rightarrow 7 p (p - 4) + 4 (p - 4) = 0 \Rightarrow (7 p + 4) (p - 4) = 0 \Rightarrow 7 p + 4 = 0 or p - 4 = 0 \Rightarrow p = - \frac{4}{7} or p = 4$

Therefore, the value of p is 4 or $- \frac{4}{7}$ .

Now, for p = 4, the equation becomes

$9 x^{2} - 30 x + 25 = 0 \Rightarrow 9 x^{2} - 15 x - 15 x + 25 = 0 \Rightarrow 3 x (3 x - 5) - 5 (3 x - 5) = 0 \Rightarrow (3 x - 5)^{2} = 0 \Rightarrow x = \frac{5}{3}, \frac{5}{3}$

for p = $- \frac{4}{7}$ , the equation becomes

$(- \frac{8}{7} + 1) x^{2} - (- 4 + 2) x + (- 4 - 3) = 0 \Rightarrow (\frac{- 8 + 7}{7}) x^{2} + 2 x - 7 = 0 \Rightarrow - \frac{1}{7} x^{2} + 2 x - 7 = 0 \Rightarrow - x^{2} + 14 x - 49 = 0 \Rightarrow x^{2} - 14 x + 49 = 0 \Rightarrow x^{2} - 7 x - 7 x + 49 = 0 \Rightarrow x (x - 7) - 7 (x - 7) = 0 \Rightarrow (x - 7)^{2} = 0 \Rightarrow x = 7, 7$

Hence, the roots of the equation are $\frac{5}{3} and 7$ .

Page No 192:

Question 8:

Find the value of p for which the quadratic equation $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0, p \neq - 1$ has equal roots. Hence, find the roots of the equation.

Disclaimer: There is a misprinting in the given question. In the question 'q' is printed instead of 9.

Answer:

The given quadratic equation $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0$ , has equal roots.

Here, $a = p + 1, b = - 6 p - 6 and c = 3 p + 27$ .

As we know that $D = b^{2} - 4 a c$

Putting the values of $a = p + 1, b = - 6 p - 6 and c = 3 p + 27$ .

$D = {[- 6 (p + 1)]}^{2} - 4 (p + 1) [3 (p + 9)] = 36 (p^{2} + 2 p + 1) - 12 (p^{2} + 10 p + 9) = 36 p^{2} - 12 p^{2} + 72 p - 120 p + 36 - 108 = 24 p^{2} - 48 p - 72$

The given equation will have real and equal roots, if D = 0

Thus, $24 p^{2} - 48 p - 72 = 0$
$\Rightarrow p^{2} - 2 p - 3 = 0 \Rightarrow p^{2} - 3 p + p - 3 = 0 \Rightarrow p (p - 3) + 1 (p - 3) = 0 \Rightarrow (p + 1) (p - 3) = 0 \Rightarrow p + 1 = 0 or p - 3 = 0 \Rightarrow p = - 1 or p = 3$

Therefore, the value of p is −1, 3.

It is given that p ≠ −1, thus p = 3 only.

Now the equation becomes

$4 x^{2} - 24 x + 36 = 0 \Rightarrow x^{2} - 6 x + 9 = 0 \Rightarrow x^{2} - 3 x - 3 x + 9 = 0 \Rightarrow x (x - 3) - 3 (x - 3) = 0 \Rightarrow (x - 3)^{2} = 0 \Rightarrow x = 3, 3$

â€‹Hence, the root of the equation is 3.

Page No 192:

Question 9:

Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.

Answer:

The given quadric equation is , and roots are real.

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if

Now factorizing of the above equation

Now according to question, the value of k is positive.

Therefore, the value of

Page No 192:

Question 10:

If the roots of the equation $(b - c) x^{2} + (c - a) x + (a - b) = 0$ are equal, then prove that 2b = a + c.

Answer:

The given quadric equation is, and roots are real

Then prove that.

Here,

As we know that

Putting the value of

As we know that

The given equation will have real roots, if

Square root both side we get

Hence

Page No 192:

Question 11:

If the roots of the equation (a² + b²)x² − 2 (ac + bd)x + (c² + d²) = 0 are equal, prove that $\frac{a}{b} = \frac{c}{d}$ .

Answer:

The given quadric equation is, and roots are real

Then prove that.

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Square root both sides we get,

Hence

Page No 192:

Question 12:

If the equation $(1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0$ has equal roots, prove that c² = a²(1 + m²).

Answer:

The given equation, has equal roots

Then prove that.

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Hence,

Page No 192:

Question 13:

Determine the nature of the roots of the following quadratic equations:

(i) (ii) , $a \neq 0, b \neq 0$

(iii) (iv)

Answer:

(i) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(ii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iv) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

Page No 192:

Question 14:

If the roots of the equations ax² + 2bx + c = 0 and $b x^{2} - 2 \sqrt{a c} x + b = 0$ are simultaneously real, then prove that b² = ac.

Answer:

The given equations are

…... (1)

…… (2)

Roots are simultaneously real

Then prove that.

Let be the discriminants of equation (1) and (2) respectively,

Then,

And

Both the given equation will have real roots, if

…… (3)

…... (4)

From equations (3) and (4) we get

Hence,

Page No 192:

Question 15:

If p, q are real and p ≠ q, then show that the roots of the equation (p − q) x² + 5(p + q) x − 2(p − q) = 0 are real and unequal.

Answer:

The quadric equation is

Here,

As we know that

Putting the value of

$D = {\{5 (p + q)\}}^{2} - 4 (p - q) (- 2 (p - q)) = 25 (p^{2} + 2 p q + q^{2}) + 8 (p^{2} - 2 p q + q^{2}) = 25 p^{2} + 50 p q + 25 q^{2} + 8 p^{2} - 16 p q + 8 q^{2} = 33 p^{2} + 34 p q + 33 q^{2}$

Since, P and q are real and , therefore, the value of .

Thus, the roots of the given equation are real and unequal.

Hence, proved

Page No 192:

Question 16:

If the roots of the equation $(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + b^{2} - a c = 0$ are equal, prove that either a = 0 or a³ + b³ + c³ = 3abc.

Answer:

The given quadric equation is, and roots are equal.

Then prove that either or

Here,

As we know that

Putting the value of

The given equation will have real roots, if

So, either

Hence

Page No 192:

Question 17:

Show that the equation $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ has no real roots, when a ≠ b.

Answer:

The quadric equation is

Here,

As we know that

Putting the value of

We have,

Thus, the value of

Therefore, the roots of the given equation are not real

Hence, proved

Page No 192:

Question 18:

Prove that both the roots of the equation $(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0$ are real but they are equal only when a = b = c.

Answer:

The quadric equation is

Here,

After simplifying the equation

As we know that

Putting the value of

Since, . So the solutions are real

Let

Then

Thus, the value of

Therefore, the roots of the given equation are real and but they are equal only when,

Hence proved

Page No 192:

Question 19:

If a, b, c are real numbers such that ac ≠ 0, then show that at least one of the equations ax² + bx + c = 0 and −ax² + bx + c = 0 has real roots.

Answer:

The given equations are

…... (1)

…… (2)

Roots are simultaneously real

Let be the discriminants of equation (1) and (2) respectively,

Then,

And

Both the given equation will have real roots, if.
Thus,
$b^{2} - 4 a c \geq 0$
$b^{2} \geq 4 a c$ ...... (3)
And,

…... (4)

Now given that are real number and as well as from equations (3) and (4) we get

At least one of the given equation has real roots

Hence, proved

Page No 199:

Question 1:

The sum of squares of two consecutive odd positive integers is 394. Find them.

Answer:

Let two consecutive odd positive integer be and other

Then according to question

Since, x being a positive number, so x cannot be negative.

Therefore,

When then odd positive

And

Thus, two consecutive odd positive integer be

Page No 199:

Question 2:

The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.

Answer:

Let first numbers be x and other

Then according to question

Thus, two consecutive number be

Page No 199:

Question 3:

Two number differ by 3 and their product is 504. Find the numbers.

Answer:

Let two required numbers be x and

Then according to question

Since, x being a number,

Therefore,

When then

And when then

Thus, two consecutive number be either

Page No 199:

Question 4:

The sum of two number a and b is 15, and the sum of their reciprocals $\frac{1}{a} and \frac{1}{b}$ is 3/10. Find the numbers a and b.

Answer:

Given that be two numbers in such a way that.

Then according to question

By cross multiplication

…. (1)

Now putting the value of b in equation (1)

Therefore,

When then

And when then

Thus, two consecutive number be either

Page No 199:

Question 5:

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.

Answer:

Let one numbers be x then other (9 − x).

Then according to question
$\frac{1}{x} + \frac{1}{(9 - x)} = \frac{1}{2} \Rightarrow \frac{(9 - x) + x}{x (9 - x)} = \frac{1}{2} \Rightarrow \frac{9}{x (9 - x)} = \frac{1}{2}$

By cross multiplication

$\Rightarrow 18 = x (9 - x) \Rightarrow x^{2} - 9 x + 18 = 0 \Rightarrow x^{2} - 6 x - 3 x + 18 = 0 \Rightarrow (x - 6) x - 3 (x - 6) = 0 \Rightarrow (x - 6) (x - 3) = 0 \Rightarrow x = 6, 3$

Since, x being a number,

Therefore,

When x = 6 then

(9 − x) = (9 − 6) = 3

When x = 3 then

(9 − x) = (9 − 3) = 6

Thus, two consecutive number be either 3, 6 or 6, 3.

Page No 199:

Question 6:

Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.

Answer:

Let three consecutive integer be and

Then according to question

Since, x being a positive number, so x cannot be negative.

Therefore,

When then other positive integer

And

Thus, three consecutive positive integer be

Page No 199:

Question 7:

The difference of squares of two number is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.

Answer:

Let the smaller numbers be x

Then according to question,

The larger number be, then

Since, x being a positive integer so, x cannot be negative,

Therefore,

When then larger number be

Thus, two consecutive number be either

Page No 199:

Question 8:

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

Answer:

Let one of the number be x then the other number be x + 2.

Then according to question,

$x^{2} + {(x + 2)}^{2} = 394 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 394 \Rightarrow 2 x^{2} + 4 x - 390 = 0 \Rightarrow x^{2} + 2 x - 195 = 0 \Rightarrow x^{2} + 15 x - 13 x - 195 = 0 \Rightarrow x (x + 15) - 13 (x + 15) = 0 \Rightarrow (x - 13) (x + 15) = 0 \Rightarrow x - 13 = 0 or x + 15 = 0 \Rightarrow x = 13 or x = - 15$

Since, x being an odd number,

Therefore, x = 13.

Then another number will be

$x + 2 = 13 + 2 = 15$

Thus, the two consecutive odd numbers are 13 and 15.

Page No 199:

Question 9:

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

Answer:

Let one of the number be 7x then the other number be 7(x + 1).

Then according to question,

${(7 x)}^{2} + {[7 (x + 1)]}^{2} = 637 \Rightarrow 49 x^{2} + 49 (x^{2} + 2 x + 1) = 637 \Rightarrow 49 x^{2} + 49 x^{2} + 98 x + 49 - 637 = 0 \Rightarrow 98 x^{2} + 98 x - 588 = 0 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow x^{2} + 3 x - 2 x - 6 = 0 \Rightarrow x (x + 3) - 2 (x + 3) = 0 \Rightarrow (x - 2) (x + 3) = 0 \Rightarrow x - 2 = 0 or x + 3 = 0 \Rightarrow x = 2 or x = - 3$

Since, the numbers are multiples of 7,

Therefore, one number = 7 × 2 =14.

Then another number will be

$7 (x + 1) = 7 \times 3 = 21$

Thus, the two consecutive multiples of 7 are 14 and 21.

Page No 199:

Question 10:

The sum of the squares of two consecutive even numbers is 340. Find the numbers.

Answer:

Let one of the number be x then the other number be x + 2.

Then according to question,

$x^{2} + {(x + 2)}^{2} = 340 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 340 \Rightarrow 2 x^{2} + 4 x - 336 = 0 \Rightarrow x^{2} + 2 x - 168 = 0 \Rightarrow x^{2} + 14 x - 12 x - 168 = 0 \Rightarrow x (x + 14) - 12 (x + 14) = 0 \Rightarrow (x - 12) (x + 14) = 0 \Rightarrow x - 12 = 0 or x + 14 = 0 \Rightarrow x = 12 or x = - 14$

Since, x being an even number,

Therefore, x = 12.

Then another number will be

$x + 2 = 12 + 2 = 14$

Thus, the two consecutive even numbers are 12 and 14.

Page No 199:

Question 11:

The sum of a number and its positive square root is 6/25. Find the number.

Answer:

Let first numbers be x

Then according to question

Let then

Since, being a positive number, so y cannot be negative.

Therefore,

Thus, the required number be

Page No 199:

Question 12:

The difference of two numbers is 4. If the difference of their reciprocal is $\frac{4}{21}$ , find the numbers.

Answer:

Let one numbers be x then other.

Then according to question

By cross multiplication

Since, x being a number,

Therefore,

When then

And when then

Thus, two consecutive number be either

Page No 199:

Question 13:

A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.

Answer:

Let the require digit be

Then according to question

.….. (1)

And, .…..(2)

Now putting the value of y in equation (2) from (1)

So, either

So, the digit can never be negative.

When then

Therefore, number

Thus, the required number be

Page No 199:

Question 14:

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.

Answer:

Let the larger number be x.

Then according to the question,

Square of the smaller number = 8x, then

$x^{2} - 8 x = 180 \Rightarrow x^{2} - 8 x - 180 = 0 \Rightarrow x^{2} - 18 x + 10 x - 180 = 0 \Rightarrow x (x - 18) + 10 (x - 18) = 0 \Rightarrow (x + 10) (x - 18) = 0 \Rightarrow x + 10 = 0 or x - 18 = 0 \Rightarrow x = - 10 or x = 18$

Since, x being a positive integer so, x cannot be negative,

Therefore, larger number = 18.

then the smaller number = $\sqrt{8 \times 18} = 12$

Thus, the two positive numbers are 12 and 18.

Page No 199:

Question 15:

The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$ . Find the numbers.

Answer:

Let the smaller number be x then the other number be 3 + x.

Then according to question,

$\frac{1}{x} - \frac{1}{3 + x} = \frac{3}{28} \Rightarrow \frac{3 + x - x}{x (3 + x)} = \frac{3}{28} \Rightarrow \frac{3}{3 x + x^{2}} = \frac{3}{28} \Rightarrow 28 = 3 x + x^{2} \Rightarrow x^{2} + 3 x - 28 = 0 \Rightarrow x^{2} + 7 x - 4 x - 28 = 0 \Rightarrow x (x + 7) - 4 (x + 7) = 0 \Rightarrow (x - 4) (x + 7) = 0 \Rightarrow x - 4 = 0 or x + 7 = 0 \Rightarrow x = 4 or x = - 7$

Since, x being a natural number,

Therefore, x = 4.

Then another number will be

$3 + x = 3 + 4 = 7$

Thus, the two natural numbers are 7 and 4.

Page No 199:

Question 16:

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$ . Find the original fraction.

Answer:

Let the denominator of the original fraction be x then the numerator be x − 3.

Then according to question,

$\frac{x - 3}{x} + \frac{x - 3 + 2}{x + 2} = \frac{29}{20} \Rightarrow \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20} \Rightarrow \frac{(x - 3) (x + 2) + (x - 1) x}{x (x + 2)} = \frac{29}{20} \Rightarrow \frac{x^{2} - 3 x + 2 x - 6 + x^{2} - x}{x^{2} + 2 x} = \frac{29}{20} \Rightarrow 20 (2 x^{2} - 2 x - 6) = 29 (x^{2} + 2 x) \Rightarrow 40 x^{2} - 40 x - 120 - 29 x^{2} - 58 x = 0 \Rightarrow 11 x^{2} - 98 x - 120 = 0 \Rightarrow 11 x^{2} - 110 x + 12 x - 120 = 0 \Rightarrow 11 x (x - 10) + 12 (x - 10) = 0 \Rightarrow (11 x + 12) (x - 10) = 0 \Rightarrow 11 x + 12 = 0 or x - 10 = 0 \Rightarrow x = - \frac{12}{11} or x = 10$

Since, x being an integer,

Therefore, x = 10.

Then the numerator will be

$x - 3 = 10 - 3 = 7$

Thus, the original fraction is $\frac{7}{10}$ .

Page No 199:

Question 17:

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Answer:

Let the given number be x.
Given that the square of a natural number diminished by 84 is equal to thrice of 8 more than the given number.
$\Rightarrow x^{2} - 84 = 3 (8 + x) \Rightarrow x^{2} - 84 = 24 + 3 x \Rightarrow x^{2} - 3 x - 108 = 0 \Rightarrow x^{2} - 12 x + 9 x - 108 = 0 \Rightarrow x (x - 12) + 9 (x - 12) = 0 \Rightarrow (x + 9) (x - 12) = 0 \Rightarrow x = - 9 or x = 12$
We ignore the negative value as we have taken natural numbers under consideration.
Hence, x = 12 is the required number.

Page No 199:

Question 18:

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

Answer:

Let the natural number be N.
Now, $(N + 12) = \frac{160}{N}$
$\Rightarrow N^{2} + 12 N - 160 = 0 \Rightarrow N (N + 20) - 8 (N + 20) = 0 \Rightarrow N = 8 (N = - 20; Not a natural number)$
Hence, required natural number is 8.

Page No 203:

Question 1:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.

Answer:

Let the original speed of train be. Then,

Increased speed of the train

Time taken by the train under usual speed to cover

Time taken by the train under increased speed to cover

Therefore,

So, either

But, the speed of the train can never be negative.

Hence, the original speed of train is

Page No 203:

Question 2:

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer:

Let the original speed of train be. Then,

Increased speed of the train

Time taken by the train under usual speed to cover

Time taken by the train under increased speed to cover

Therefore,

So, either

But, the speed of the train can never be negative.

Hence, the original speed of train is

Page No 203:

Question 3:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of two trains.

Answer:

Let the speed of the passenger train be. Then,

Speed of the express train

Time taken by the passenger train to cover between Mysore to Bangalore

Time taken by the express train to cover between Mysore to Bangalore

Therefore,

So, either

But, the speed of the train can never be negative.

Thus, when then speed of express train

Hence, the speed of the passenger train is

and the speed of the express train is respectively.

Page No 203:

Question 4:

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream that to return down stream to the same spot. Find the speed of the stream.

Answer:

Let the speed of the stream be x km/hr.

speed of the boat in still water = 18 km/hr.
Total Distance = 24 km.

We know that,
Speed of the boat up stream = speed of the boat in still water − speed of the stream
= (18 − x) km/hr

Speed of the boat down stream = speed of the boat in still water + speed of the stream
= (18 + x) km/hr

Time of up stream journey = t₁ = $\frac{24}{18 - x}$ hr

Time of down stream journey = t₂ = $\frac{24}{18 + x}$ hr

According to the question,

t₁ − t₂ = 1 hr
$\Rightarrow \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \Rightarrow \frac{24 (18 + x - 18 + x)}{(18 - x) (18 + x)} = 1 \Rightarrow \frac{24 (2 x)}{(18)^{2} - x^{2}} = 1 \Rightarrow 48 x = 324 - x^{2} \Rightarrow x^{2} + 48 x - 324 = 0 \Rightarrow x^{2} + 54 x - 6 x - 324 = 0 \Rightarrow x (x + 54) - 6 (x + 54) = 0 \Rightarrow (x - 6) (x + 54) = 0 \Rightarrow x - 6 = 0 or x + 54 = 0 \Rightarrow x = 6 or x = - 54$

Since, speed cannot be negative.

Thus, speed of the stream is 6 km/hr.

Page No 203:

Question 5:

A motor boat whose speed in still water is 9 km/hr, goes 15 km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.

Answer:

Let speed of stream be x km/h.

Given:

Speed of boat = 9 km/h
Distance covered upstream = 15 km
Distance covered downstream = 15 km
Total time taken = 3 hours 45 minutes = $\frac{15}{4} hours$

Now, Speed of boat upstream = 9 − x km/h
Speed of boat downstream = 9 + x km/h

$\frac{Distance}{Speed} = Time$

According to the question,

$\frac{15}{9 + x} + \frac{15}{9 - x} = \frac{15}{4} \Rightarrow \frac{15 (9 - x) + 15 (9 + x)}{(9 + x) (9 - x)} = \frac{15}{4} \Rightarrow \frac{135 - 15 x + 135 + 15 x}{81 - x^{2}} = \frac{15}{4} \Rightarrow \frac{135 + 135}{81 - x^{2}} = \frac{15}{4} \Rightarrow \frac{270}{81 - x^{2}} = \frac{15}{4} \Rightarrow \frac{18}{81 - x^{2}} = \frac{1}{4} \Rightarrow 18 (4) = 81 - x^{2} \Rightarrow 72 = 81 - x^{2} \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3 But x is the speed of stream which is always positive . Thus, x = 3 km / h$

Hence, the speed of the stream is 3 km/h.

Page No 203:

Question 6:

A train , travelling at a uniform speed for 360 km , would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train .

Answer:

Let the speed of train be x km/h

Distance to be travelled = 360 km

We know that,

Time take by the train intially when it was travelling with uniform speed of x km/h =

If the speed was increased by 5 km/h

Time taken by train = $fraction numerator 360 over denominator x plus 5 end fraction$

With increased speed, the time taken is 48 min less. So, difference in time will be

$360 over x minus fraction numerator 360 over denominator x plus 5 end fraction equals 48 over 60rightwards double arrow 360 open square brackets fraction numerator x plus 5 minus x over denominator x open parentheses x plus 5 close parentheses end fraction close square brackets equals 4 over 5rightwards double arrow fraction numerator 5 over denominator x open parentheses x plus 5 close parentheses end fraction equals 1 over 450rightwards double arrow x squared plus 5 x minus 2250 equals 0rightwards double arrow open parentheses x plus 50 close parentheses open parentheses x minus 45 close parentheses equals 0rightwards double arrow x equals negative 50 comma space 45$
Hence, the speed of train = 45 km/hr

Page No 203:

Question 7:

A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

Answer:

Let the usual speed of plane be. Then,

Increased speed of the plane

Time taken by the plane under usual speed to cover

Time taken by the plane under increased speed to cover

Therefore,

So, either

But, the speed of the plane can never be negative.

Hence, the usual speed of train is

Page No 204:

Question 8:

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed . If it takes 3 hours to complete total journey , what is its original average speed ?

Answer:

Let the original speed of the train be x.
For a distance of 63 km, let the speed be x km/h.
So, time = $\frac{63}{x}$
For a distance of 72 km, speed = 6 + x km/h
Time = $\frac{72}{x + 6}$
Total time = 3 hours
$\frac{63}{x} + \frac{72}{x + 6} = 3 \Rightarrow \frac{63 x + 378 + 72 x}{x (x + 6)} = 3 \Rightarrow 63 x + 378 + 72 x = 3 x^{2} + 18 x \Rightarrow 3 x^{2} - 117 x - 378 = 0 \Rightarrow (3 x - 126) (x + 3) = 0 \Rightarrow x = - 3, 42$

So, the original speed = 42 km/h.

Page No 204:

Question 9:

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Answer:

Let the usual speed of aero plane be. Then,

Increased speed of the aero plane

Time taken by the aero plane under usual speed to cover

Time taken by the aero plane under increased speed to cover

Therefore,

So, either

But, the speed of the aero plane can never be negative.

Hence, the usual speed of train is

Page No 204:

Question 10:

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

Answer:

Let the original speed of the plane be x km/hr.

Increased speed of the plane = (x + 100) km/hr.
Total Distance = 1500 km.

We know that, $Time = \frac{Distance}{Speed}$

Time taken to reach the destination at original speed = t₁ = $\frac{1500}{x}$ hr

Time taken to reach the destination at increasing speed = t₂ = $\frac{1500}{x + 100}$ hr

According to the question,

t₁ − t₂ = 30 min
$\Rightarrow \frac{1500}{x} - \frac{1500}{x + 100} = \frac{30}{60} \Rightarrow \frac{1500 (x + 100) - 1500 x}{x (x + 100)} = \frac{1}{2} \Rightarrow \frac{1500 x + 150000 - 1500 x}{x^{2} + 100 x} = \frac{1}{2} \Rightarrow \frac{150000}{x^{2} + 100 x} = \frac{1}{2} \Rightarrow 300000 = x^{2} + 100 x \Rightarrow x^{2} + 100 x - 300000 = 0 \Rightarrow x^{2} + 600 x - 500 x - 300000 = 0 \Rightarrow x (x + 600) - 500 (x + 600) = 0 \Rightarrow (x - 500) (x + 600) = 0 \Rightarrow x - 500 = 0 or x + 600 = 0 \Rightarrow x = 500 or x = - 600$

Since, speed cannot be negative.

Thus, the original speed/hour of the plane is 500 km/hr.

Page No 204:

Question 11:

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.

Answer:

Distance covered = 2592 km
Let the speed of the car be x km/h
According to the question, Time = $\frac{speed}{2} = \frac{x}{2} h$
we know that
$speed = \frac{distance}{time}$
$x = \frac{2592}{x / 2} x^{2} = 5184 x = \sqrt{5184} x = 72 Speed of the car = 72 km / h Time taken to cover the given distance = 36 h$

Page No 205:

Question 1:

Ashu is x years old while his mother Mrs Veena is x² years old. Five years hence Mrs Veena will be three times old as Ashu. Find their present ages.

Answer:

Given that Ashu’s present age isand his mother Mrs. Veena is

Then according to question,

Five years later, Ashu’s is

And his mother Mrs. Veena is

Thus

So, either

But, the age can never be negative.

Therefore, when then

Hence, Ashu’s present age isand his mother Mrs. Veena is

Page No 205:

Question 2:

The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.

Answer:

Let the present age of the man be

Then present age of his son is

Five years ago, man’s age

And his son’s age

Then according to question,

So, either

But, the father’s age never be 5 years

Therefore, when then

Hence, man’s present age isand his son’s age is

Page No 205:

Question 3:

The product of Shikha's age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.

Answer:

Let the present age of Shikha be

Then, 8 years later, age of her

Five years ago, her age

Then according to question,

So, either

But the age never be negative

Hence, the present age of Shikha be

Page No 205:

Question 4:

The product of Ramu's age (in years) five years ago and his age (in years) nice years later is 15. Determine Ramu's present age.

Answer:

Let the present age of Ramu be

Then, 9 years later, age of her

Five years ago, her age

Then according to question,

So, either

But the age never be negative

Hence, the present age of Ramu be

Page No 205:

Question 5:

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years ago, the product of their ages in years was 48.

Answer:

Let the present age of two friends be x years and (20 − x) years respectively.

Then, 4 years later, the age of two friends will be (x − 4) years and (20 − x − 4) years respectively

Then according to question,

$- x^{2} + 20 x - 64 - 48 = 0$
$x^{2} - 20 x + 112 = 0$

Let D be the discriminant of the above quadratic equation.

Then,

Putting the value of a = 1, b = − 20 and c = 112
$D = {(- 20)}^{2} - 4 \times 1 \times 112$
= 400 − 448
= − 48

Thus,

So, the above equation does not have real roots.

Hence, the given situation is not possible.

Page No 205:

Question 6:

A girls is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.

Answer:

Let the present age of girl bethen, age of her sister

Then, 4 years later, age of girl and her sister’s age be

Then according to question,

So, either

But the age never be negative

Therefore, when then

Hence, the present age of girl be and her sister’s age be

Page No 205:

Question 7:

The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer:

Let the present age of Rehman be

Then, 8 years later, age of her

Five years ago, her age

Then according to question,

So, either

But the age never be negative

Hence, the present age of Rehman be

Page No 206:

Question 8:

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?

Answer:

Let, the present age of Zeba be x years.
If she were 5 years younger, then the square of her age would have been 11 more than 5 times her actual age
$\Rightarrow {(x - 5)}^{2} = 11 + 5 x \Rightarrow x^{2} + 25 - 10 x = 11 + 5 x \Rightarrow x^{2} - 15 x + 14 = 0 \Rightarrow x^{2} - 14 x - x + 14 = 0 \Rightarrow x (x - 14) - 1 (x - 14) = 0 \Rightarrow (x - 14) (x - 1) = 0 \Rightarrow x = 14 or x = 1$
The age cannot be 1 year as we have also talked about the age 5 years ago.
Therefore, Zeba's present age is $x = 14$ years.

Page No 206:

Question 9:

At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Answer:

Let, the present age of Asha be a years
Present age of Nisha be n years.
Given that at present Asha's age is 2 more than the square of her daughter Nisha's age.
$\Rightarrow a = 2 + n^{2}$
So, we can say that Nisha will take $(2 + n^{2}) - n$ years to reach her mother's age.
So, after $(2 + n^{2}) - n$ years, her mother's age will be $(2 + n^{2}) + (2 + n^{2} - n)$ years.
It is also given that when Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha.
Therefore, after $2 + n^{2} - n$ years, Asha's age will be $10 n - 1$ years
$\Rightarrow 2 + n^{2} + 2 + n^{2} - n = 10 n - 1 \Rightarrow 2 n^{2} - 10 n - n + 5 = 0 \Rightarrow 2 n (n - 5) - 1 (n - 5) = 0 \Rightarrow (2 n - 1) (n - 5) = 0 \Rightarrow n = \frac{1}{2} or n = 5$
We ignore the fractional value.
Therefore, Nisha's age is n = 5 years.
Asha's age = $2 + n^{2} = 2 + 25 = 27$ years

Page No 207:

Question 1:

The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.

Answer:

Let the length of one side of right triangle be then other side be

And given that hypotenuse

As we know that by Pythagoras theorem,

So, either

But the side of right triangle can never be negative

Therefore, when then

Hence, length of one side of right triangle be then other side be

Page No 207:

Question 2:

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Answer:

Let the length of smaller side of rectangle be then larger side be and their diagonal be

Then, as we know that Pythagoras theorem

But, the side of rectangle can never be negative.

Therefore, when then

Hence, length of smaller side of rectangle be and larger side be

Page No 207:

Question 3:

The hypotenuse of a right triangle is $3 \sqrt{10} cm$ . If the smaller leg is tripled and the longer leg doubled, new hypotenuse wll be $9 \sqrt{5} cm$ . How long are the legs of the triangle?

Answer:

Let the length of smaller side of right triangle be then larger side be

Then, as we know that by Pythagoras theorem

….. (1)

If the smaller side is triple and the larger side be doubled, the new hypotenuse is

Therefore,

….. (2)

From equation (1) we get

Now putting the value of in equation (2)

But, the side of right triangle can never be negative

Therefore, when then

Hence, length of smaller side of right triangle be then larger side be

Page No 207:

Question 4:

A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answer:

Let P be the required location on the boundary of a circular park such that its distance from gate B is that is BP

Then, AP

In the right triangle ABP we have by using Pythagoras theorem

But the side of right triangle can never be negative

Therefore,

Hence, P is at a distance of from the gate B.

Page No 212:

Question 1:

Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²? If so, find its length and breadth.

Answer:

Let the breadth of the rectangular mango grove be x meter and the length. Then

Area of the rectangle

Sides of the rectangular hall never be negative

Therefore, length

, it is possible.

Hence, breadth of the hall be be

Page No 212:

Question 2:

Is it possible to design a rectangular park of perimeter 80 m and area 400 m². If so, find its length and breadth.

Answer:

Let the breadth of the rectangle be. Then

Perimeter

And area of the rectangle

, it is possible.

Hence, breadth of the rectangular park be and length be

Page No 212:

Question 3:

Sum of the areas of two squares is 640 m². If the difference of their perimeters is 64 m. Find the sides of the two squares.

Answer:

Let the sides of the squares are x m and .Then

According to question,

Sum of the difference of their perimeter=64 m

….. (1)

And sum of the areas of square

….. (2)

Putting the value of x in equation (2) from equation (1)

Sides of the square never are negative.

Therefore, putting the value of x in equation (1)

Hence, sides of the square be and respectively.

Page No 212:

Question 4:

The area of a rectangular plot is 528 m². The length of the plot (in metres) is one metre more then twice its breadth. Find the length and the breadth of the plot.

Answer:

Let the breadth of the rectangular plot be x m.
Then, the length of the rectangular plot = (1 + 2x) m.

According to the question,

Length × Breadth = Area
$x (1 + 2 x) = 528 \Rightarrow 2 x^{2} + x - 528 = 0 \Rightarrow 2 x^{2} + 33 x - 32 x - 528 = 0 \Rightarrow x (2 x + 33) - 16 (2 x + 33) = 0 \Rightarrow (x - 16) (2 x + 33) = 0 \Rightarrow x - 16 = 0 or 2 x + 33 = 0 \Rightarrow x = 16 or x = - \frac{33}{2}$

Since, length and breadth of the rectangle cannot be negative.

Thus, the breadth of the rectangular plot is 16 m.

and the length of the rectangular plot is (1 + 2×16) = 33 m.

Page No 212:

Question 5:

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m². Find the length and breadth of the pond.

Answer:

It is given that the dimensions of rectangular park is 50 m × 40 m.
∴ Area of the rectangular park = 50 × 40 = 2000 m²
Area of the grass surrounding the pond = 1184 m²
Now,
Area of the rectangular pond
= Area of the rectangular park − Area of the grass surrounding the rectangular pond
= 2000 − 1184
= 816 m²
Let the uniform width of the surrounding grass be x.
∴ Length of the rectangular pond = (50 − 2x) m
Breadth of the rectangular pond = (40 − 2x) m
Now,
Area of rectangular pond = 816 m²
∴ (50 − 2x) × (40 − 2x) = 816
⇒ 2000 − 80x − 100x + 4x² = 816
⇒ 4x² − 180x + 2000 − 816 = 0
⇒ 4x² − 180x + 1184 = 0
⇒ x² − 45x + 296 = 0
⇒ x² − 37x − 8x + 296 = 0
⇒ x(x − 37) − 8(x − 37) = 0
⇒ (x − 8)(x − 37) = 0
⇒ x − 8 = 0 or x − 37 = 0
⇒ x = 8 or x = 37
For x = 37,
Length of rectangular pond = 50 − 2 × 37 = −24 m, which is not possible
So, x ≠ 37
Therefore, x = 8.
When x = 8,
Length of the rectangular pond = 50 − 2 × 8 = 50 − 16 = 34 m
Breadth of the rectangular pond = 40 − 2 × 8 = 40 − 16 = 24 m.

Page No 213:

Question 1:

A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

Answer:

Let B alone takes x days to finish the work. Then, B’s one day’s work.

Similarly, A alone can finish it in days to finish the work. Then, A’s one day’s work.

It is given that

A’s one day’s work + B’s one day’s work’s one day’s work

But is not correct.

therefore, is correct

Hence, the time taken by B to finish the work in

Page No 214:

Question 2:

If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?

Answer:

Let the first pipe takes x hours to fill the reservoir. Then the second pipe will takes hours to fill the reservoir.

Since, the faster pipe takes x hours to fill the reservoir.

Therefore, portion of the reservoir filled by the faster pipe in one hour

So, portion of the reservoir filled by the faster pipe in 12 hours

Similarly,

Portion of the reservoir filled by the slower pipe in 12 hours

It is given that the reservoir is filled in 12 hours.

So,

But, cannot be negative.

Therefore, when then

Hence, the second pipe will takes to fill the reservoir.

Page No 214:

Question 3:

Two water taps together can fill a tank in $9 \frac{3}{8} hours$ . The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer:

Let the first water tape takes x hours to fill the tank. Then the second water tape will takes hours to fill the tank.

Since, the faster water tape takes x hours to fill the tank.

Therefore, portion of the tank filled by the faster water tape in one hour

So, portion of the tank filled by the faster water tape in hours

Similarly,

Portion of the tank filled by the slower water tape in hours

It is given that the tank is filled in hours.

So,

But, cannot be negative.

Therefore, when then

Hence, the first water tape will takes to fill the tank, and the second water tape will takes to fill the tank.

Page No 214:

Question 4:

Two pipes running together can fill a tank in $11 \frac{1}{9} minutes$ . If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Answer:

Let the first pipe takes x minutes to fill the tank. Then the second pipe will takes minutes to fill the tank.

Since, the first pipe takes x minutes to fill the tank.

Therefore, portion of the tank filled by the first pipe in one minutes

So, portion of the tank filled by the first pipe in minutes

Similarly,

Portion of the tank filled by the second pipe in minutes

It is given that the tank is filled in minutes.

So,

But, cannot be negative.

Therefore, when x = 20 then

Hence, the first water tape will takes 20 min to fill the tank, and the second water tape will take 25 min to fill the tank.

Page No 214:

Question 5:

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Answer:

Let the pipe of larger diameter takes x hours.
Then, the pipe of smaller diameter takes x + 10 hours to fill the pool.

Now, the part of the pool filled by the larger pipe in 1 hour = $\frac{1}{x}$
and the part of the pool filled by the smaller pipe in 1 hour = $\frac{1}{x + 10}$

If the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, only half of the pool can be filled,

$∴ \frac{4}{x} + \frac{9}{x + 10} = \frac{1}{2} \Rightarrow \frac{4 (x + 10) + 9 x}{x (x + 10)} = \frac{1}{2} \Rightarrow \frac{4 x + 40 + 9 x}{x^{2} + 10 x} = \frac{1}{2} \Rightarrow 2 (13 x + 40) = x^{2} + 10 x \Rightarrow x^{2} + 10 x - 26 x - 80 = 0 \Rightarrow x^{2} - 16 x - 80 = 0 \Rightarrow x^{2} - 20 x + 4 x - 80 = 0 \Rightarrow x (x - 20) + 4 (x - 20) = 0 \Rightarrow (x + 4) (x - 20) = 0 \Rightarrow x + 4 = 0 or x - 20 = 0 \Rightarrow x = - 4 or x = 20$

Since, time cannot be negative.

∴ x = 20

Thus, the pipe of larger diameter takes 20 hours and the pipe of smaller diameter takes (20 + 10) = 30 hours to fill the pool separately.

Page No 214:

Question 6:

Two water taps together can fill a tank in $1 \frac{7}{8}$ hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer:

Let the smaller tap takes x hours to fill the tank.
Then, the larger one takes (x − 2) hours to fill the tank.

Tank filled in 1 hour by smaller tap = $\frac{1}{x}$
Tank filled in 1 hour by larger tap = $\frac{1}{x - 2}$
Tank filled in 1 hour by both the taps = $\frac{8}{15}$

According to the question,

$\frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \Rightarrow \frac{(x - 2) + (x)}{(x) (x - 2)} = \frac{8}{15} \Rightarrow \frac{2 x - 2}{x^{2} - 2 x} = \frac{8}{15} \Rightarrow 15 (2 x - 2) = 8 (x^{2} - 2 x) \Rightarrow 30 x - 30 = 8 x^{2} - 16 x \Rightarrow 8 x^{2} - 46 x + 30 = 0 \Rightarrow 4 x^{2} - 23 x + 15 = 0 \Rightarrow x = \frac{- (- 23) \pm \sqrt{{(- 23)}^{2} - 4 (4) (15)}}{2 (4)} \Rightarrow x = \frac{23 \pm \sqrt{529 - 240}}{8} \Rightarrow x = \frac{23 \pm \sqrt{289}}{8} \Rightarrow x = \frac{23 \pm 17}{8} \Rightarrow x = \frac{23 + 17}{8} or x = \frac{23 - 17}{8} \Rightarrow x = 5 or x = 0.75 But x \neq 0.75 (∵ (x - 2) becomes negative) Thus, x = 5$

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.

Page No 217:

Question 1:

A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter costs Rs. one less, the cost would remain unchanged. How long is the piece?

Answer:

Let the length of the piece be x metres.

Then, rate per metre

According to question, new length.

Since the cost remain same. Therefore, new rate per metre

It is given that

Because x cannot be negative.

Thus, is the require solution.

Therefore, the length of the piece be

Page No 217:

Question 2:

Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic?

Answer:

Let x students planned a picnic.

Then, the share of each student

According to question, 8 students fail to go picnic, then remaining students.

Therefore, new share of each student

It is given that

Because x cannot be negative.

Thus, the total numbers of students attend a picnic

Therefore, the total numbers of students attend a picnic be

Page No 217:

Question 3:

If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy.

Answer:

Let the original list price of the toy be Rs. x .

Then, the number of toys brought for Rs.360

According to question, reduced list price of the toys.

Therefore, the number of toys brought for Rs.360

It is given that

Because x cannot be negative.

Thus, is the require solution.

Therefore, the original list price of the toy be

Page No 217:

Question 4:

Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.

Answer:

Let the original number of persons be x.
Then, by the given information,

Thus, the original number of persons is Rs 25.

Page No 217:

Question 5:

In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects.

Answer:

Let marks obtained by Shefali in mathematics be x, then in english

It is given that,

Therefore, when then

Hence, marks in mathematics and marks in science .

Or,

when then

Hence, marks in mathematics and marks in science .

Page No 217:

Question 6:

A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

Answer:

Let the cost price of article be Rs. x.

Then, gain percent = x

Therefore, the selling price of article

It is given that

Because x cannot be negative.

Thus, is the require solution.

Therefore, the cost price of article be

Page No 217:

Question 7:

A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distance from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answer:

Let P be the required location on the boundary of a circular park such that its distance from gate B is that is BP

Then, AP

In the right triangle ABP we have by using Pythagoras theorem

But, the side of right triangle can never be negative.

Therefore,

Hence, P is at a distance of from the gate B.

Page No 218:

Question 8:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production on each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

Answer:

Let the number of article produced by the cottage industry be x.

Then the cost of production of each article

It is given that total cost of production

Therefore,

Therefore, x cannot be negative.

So, when then

Hence, the number of article produced by the cottage industry be and the cost of production of each article .

Page No 218:

Question 9:

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{t^{2}}{4}$ minutes. Find t.

Answer:

It is given that at t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{t^{2}}{4}$ minutes i.e. $\frac{t^{2}}{4} - 3$ .
There are 60 minutes in 1 hour, so at t minutes past 2, time left to 3 will be $60 - t$ minutes.
$\Rightarrow 60 - t = \frac{t^{2}}{4} - 3 \Rightarrow 240 - 4 t = t^{2} - 12 \Rightarrow t^{2} + 4 t - 252 = 0 \Rightarrow t^{2} + 18 t - 14 t - 252 = 0 \Rightarrow t (t + 18) - 14 (t + 18) = 0 \Rightarrow (t - 14) (t + 18) = 0 \Rightarrow t = 14 or t = - 18$
We ignore the negative value because time cannot be negative.
Therefore, t = 14

Page No 218:

Question 1:

Write the value of k for which the quadratic equation x² − kx + 4 = 0 has equal roots.

Answer:

The given quadric equation is , and roots are equal.

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

Therefore, the value of

Page No 218:

Question 2:

What is the nature of roots of the quadratic equation 4x² − 12x − 9 = 0?

Answer:

The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

Page No 218:

Question 3:

Write a quadratic polynomial, sum of whose zeros is $2 \sqrt{3}$ and their product is 2.

Answer:

As we know that the quadratic polynomial

According to question,

And

Thus putting the value in above,

where k is real number.

Therefore, the quadratic polynomial be

Page No 218:

Question 4:

Show that x = −3 is a solution of x² + 6x + 9 = 0.

Answer:

Given that the equation

Therefore, is the solution of given equation.

Hence, proved.

Page No 218:

Question 5:

Show that x = −2 is a solution of 3x² + 13x + 14 = 0.

Answer:

Given that the equation

Therefore, is the solution of given equation.

Hence, proved.

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Question 6:

Find the discriminant of the quadratic equation $3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0$ .

Answer:

Given that quadric equation is .

Then, find the value of discrimenant.

Here,

As we know that discrimenant

Putting the value of

Thus, the value of discrimenant be .

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Question 7:

If $x = - \frac{1}{2}$ , is a solution of the quadratic equation $3 x^{2} + 2 k x - 3 = 0$ , find the value of k.

Answer:

Since, $x = - \frac{1}{2}$ , is a solution of the quadratic equation $3 x^{2} + 2 k x - 3 = 0$ .
So, it satisfies the given equation.

$∴ 3 {(- \frac{1}{2})}^{2} + 2 k (- \frac{1}{2}) - 3 = 0 \Rightarrow \frac{3}{4} - k - 3 = 0 \Rightarrow k = \frac{3}{4} - 3 \Rightarrow k = \frac{3 - 12}{4} \Rightarrow k = - \frac{9}{4}$

Thus, the value of k is $- \frac{9}{4}$ .

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Question 8:

If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.

Answer:

If x = 3 is a root of the quadratic equation x² − 2kx â€‹− 6 = 0, then it will satisfy the given equation.
Put x = 3 in given quadratic equation
⇒ 3² − 2k(3) − 6 = 0
⇒ k = $\frac{1}{2}$

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Question 9:

For what value of k, the roots of the equation x² + 4x + k = 0 are real?

Answer:

Let x² + 4x + k = 0 be a quadratic equation.

It is given that, it has real roots.

$\Rightarrow Discriminant \geq 0 \Rightarrow b^{2} - 4 a c \geq 0 \Rightarrow {(4)}^{2} - 4 (1) (k) \geq 0 \Rightarrow 16 - 4 k \geq 0 \Rightarrow 16 \geq 4 k \Rightarrow 4 k \leq 16 \Rightarrow k \leq 4$

Hence, the values of k must be less than or equal to 4.

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Question 10:

Find the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other.

Answer:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 3x²– 10x + k = 0.

Product of roots = $\frac{k}{3}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{3} \Rightarrow 1 = \frac{k}{3} \Rightarrow k = 3$

Hence, the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other is 3.

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Question 11:

Write the set of value of 'a' for which the equation x² + ax − 1 = 0 has real roots.

Answer:

The given quadric equation is

Then find the value of a.

Here,

As we know that

Putting the value of

The given equation will have real roots, if_.

Therefore, for all real values of a, the given equation has real roots.

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Question 12:

Is there any real value of 'a' for which the equation x² + 2x + (a² + 1) = 0 has real roots?

Answer:

Let quadratic equation has real roots.

Here,

As we know that

Putting the value of , we get
$D = {(2)}^{2} - 4 \times 1 \times (a^{2} + 1) = 4 - 4 (a^{2} + 1) = - 4 a^{2}$

The given equation will have equal roots, if

i.e. $- 4 a^{2} > 0 \Rightarrow a^{2} < 0$

which is not possible, as the square of any number is always positive.

Thus, there is no any real value of a for which the given equation has real roots.

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Question 13:

Write the set of value of k for which the quadratic equations has 2x² + kx − 8 = 0 has real roots.

Answer:

The given quadric equation is, and roots are real.

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation will have real roots, if

I.e.,

Therefore, for all real values of k, the given equation has real roots.

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Question 14:

If $1 + \sqrt{2}$ is a root of a quadratic equation will rational coefficients, write its other root.

Answer:

Given that is a root of the quadratic equation with rational coefficients.

Then find the other root.

As we know that if is a root of the quadratic equation with rational coefficients then other roots be .

Hence, the require root of the quadratic equation be

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Question 15:

Write the number of real roots of the equation x² + 3 |x| + 2 = 0.

Answer:

The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, roots of the given equation are.

∴The number of real roots of the given equation is 4.

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Question 16:

Write the sum of real roots of the equation x² + |x| − 6 = 0.

Answer:

The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

Thus, sum of the roots be

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Question 17:

Write the value of λ for which x² + 4x + λ is a perfect square.

Answer:

The given quadric equation is

Then find the value of k.

Here,

As we know that

Putting the value of

The given equation are perfect square, if

Therefore, the value of

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Question 18:

Write the condition to be satisfied for which equations ax² + 2bx + c = 0 and $b x^{2} - 2 \sqrt{a c} x + b = 0$ have equal roots.

Answer:

The given equations are

…... (1)

And, …… (2)

roots are equal.

Let be the discriminants of equation (1) and (2) respectively,

Then,

And

Both the given equation will have real roots, if

…… (3)

…... (4)

From equations (3) and (4) we get

Hence, is the condition under which the given equations have equal roots.

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Question 1:

If $(k + 1) x^{2} + \frac{3}{2} x = 7$ is a quadratic equation, then k cannot be equal to _________.

Answer:

If $a x^{2} + b x + c = 0$ is a quadratic equation, then a cannot be equal to zero.

Therefore, if $(k + 1) x^{2} + \frac{3}{2} x = 7$ is a quadratic equation, then (k + 1) cannot be equal to zero.
⇒ k ≠ −1

Hence, k cannot be equal to −1.

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Question 2:

Every quadratic equation has exactly ________ root.

Answer:

Let $a x^{2} + b x + c = 0$ be a quadratic equation.

Number of roots of any equation = Highest power of x
= 2

Hence, every quadratic equation has exactly 2 roots.

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Question 3:

The values of k for which the quadratic equation 2x² – kx + k = 0 has equal roots are ______.

Answer:

Let 2x² – kx + k = 0 be a quadratic equation.

It is given that, it has equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(- k)}^{2} - 4 (2) (k) = 0 \Rightarrow k^{2} - 8 k = 0 \Rightarrow k (k - 8) = 0 \Rightarrow k = 0, 8$

Hence, the values of k for which the quadratic equation 2x² – kx + k = 0 has equal roots are 0 or 8.

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Question 4:

If $\frac{1}{2}$ is a root of the equation $x^{2} + k x - \frac{5}{4} = 0$ , then the value of k is ________.

Answer:

Let f(x) = $x^{2} + k x - \frac{5}{4}$

It is given that, $\frac{1}{2}$ is a root of the equation $x^{2} + k x - \frac{5}{4} = 0$ .

Therefore, $f (x) = 0 when x = \frac{1}{2}$ .

On putting x = $\frac{1}{2}$ in f(x) = 0, we get

${(\frac{1}{2})}^{2} + k (\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} + k (\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1 + 2 k - 5}{4} = 0 \Rightarrow 2 k - 4 = 0 \Rightarrow 2 k = 4 \Rightarrow k = 2$

Hence, if $\frac{1}{2}$ is a root of the equation $x^{2} + k x - \frac{5}{4} = 0$ , then the value of k is 2.

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Question 5:

If one root of the quadratic equation 3x²– 10x + k = 0 is reciprocal of the other, then k = ________.

Answer:

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Question 6:

The values of k for which the quadratic equation x² – 4kx + k = 0 has equal roots, are _________.

Answer:

Let x² – 4kx + k = 0 be a quadratic equation.

It is given that, it has equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(- 4 k)}^{2} - 4 (1) (k) = 0 \Rightarrow 16 k^{2} - 4 k = 0 \Rightarrow 4 k (4 k - 1) = 0 \Rightarrow 4 k = 0 or 4 k - 1 = 0 \Rightarrow k = 0 or k = \frac{1}{4}$

Hence, the values of k for which the quadratic equation x² – 4kx + k = 0 has equal roots, are $0 or \frac{1}{4}$ .

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Question 7:

If the arithmetic mean of the roots of the equation x² – 6x + 8 = 0 is ________.

Answer:

Let the roots of the quadratic equation x² – 6x + 8 = 0 be $α and β$ .

$Sum of the roots = - \frac{- 6}{1} \Rightarrow α + β = 6 Arithmetic Mean of the roots = \frac{α + β}{2} = \frac{6}{2} = 3$

Hence, the arithmetic mean of the roots of the equation x² – 6x + 8 = 0 is 3.

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Question 8:

If the arithmetic mean of the roots of the equation x (x – 2) + 4ax = 5 is 3, then a = ________.

Answer:

Let the roots of the quadratic equation x (x – 2) + 4ax = 5 be $α and β$ .

$x (x - 2) + 4 a x = 5 \Rightarrow x^{2} - 2 x + 4 a x - 5 = 0 \Rightarrow x^{2} + x (4 a - 2) - 5 = 0 Sum of the roots = - \frac{4 a - 2}{1} \Rightarrow α + β = 2 - 4 a Arithmetic Mean of the roots = 3 \Rightarrow \frac{α + β}{2} = 3 \Rightarrow \frac{2 - 4 a}{2} = 3 \Rightarrow 2 - 4 a = 6 \Rightarrow - 4 a = 6 - 2 \Rightarrow - 4 a = 4 \Rightarrow a = - 1$

Hence, a = –1.

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Question 9:

If the equation mx² + 2x + m = 0 is satisfied by only one real value of x, then the values of m are ______ and ______.

Answer:

Let mx² + 2x + m = 0 be a quadratic equation.

Since, the equation is satisfied by only one real value of x
Therefore, the roots are real and equal.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(2)}^{2} - 4 (m) (m) = 0 \Rightarrow 4 - 4 m^{2} = 0 \Rightarrow 4 m^{2} = 4 \Rightarrow m^{2} = 1 \Rightarrow m = \pm 1$

Hence, the values of m are 1 and −1.

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Question 10:

The quadratic equation with rational coefficients having $\frac{\sqrt{3}}{2}$ as a root is _______.

Answer:

Irrational roots of a quadratic equation always occurs in pairs.

If one root is $\frac{\sqrt{3}}{2}$ then, other root is – $\frac{\sqrt{3}}{2}$ .

Now,

Quadratic equation with given roots is
x² – (sum of roots)x + (product of roots) = 0

$\Rightarrow x^{2} - (\frac{\sqrt{3}}{2} + (- \frac{\sqrt{3}}{2})) x + (\frac{\sqrt{3}}{2} \times - \frac{\sqrt{3}}{2}) = 0 \Rightarrow x^{2} + (\frac{- 3}{4}) = 0 \Rightarrow x^{2} - \frac{3}{4} = 0$

Hence, quadratic equation with rational coefficients having $\frac{\sqrt{3}}{2}$ as a root is $x^{2} - \frac{3}{4} = 0 .$

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Question 11:

The total number of values of x satisfying $x^{2} - 3 |x| + 2 = 0$ is ________.

Answer:

Given: $x^{2} - 3 |x| + 2 = 0$

Since,
$|x| = \{\begin{cases} x, if x \geq 0 \\ - x, if x < 0 \end{cases} Thus, x^{2} - 3 |x| + 2 = 0 can be written as \{\begin{cases} x^{2} - 3 x + 2 = 0, if x \geq 0 . . . (1) \\ x^{2} + 3 x + 2 = 0, if x < 0 . . . (2) \end{cases} Solving (1), we get if x \geq 0 . x^{2} - 3 x + 2 = 0 \Rightarrow x^{2} - 2 x - x + 2 = 0 \Rightarrow x (x - 2) - 1 (x - 2) = 0 \Rightarrow (x - 1) (x - 2) = 0 \Rightarrow x = 1, 2 Solving (2), we get if x < 0 . x^{2} + 3 x + 2 = 0 \Rightarrow x^{2} + 2 x + x + 2 = 0 \Rightarrow x (x + 2) + 1 (x + 2) = 0 \Rightarrow (x + 1) (x + 2) = 0 \Rightarrow x = - 1, - 2$

Hence, the total number of values of x satisfying $x^{2} - 3 |x| + 2 = 0$ is 4.

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Question 12:

The number of real roots of the equation $x^{2} + 3 |x| + 2 = 0$ is ________.

Answer:

Given: $x^{2} + 3 |x| + 2 = 0$

Since,
$|x| = \{\begin{cases} x, if x \geq 0 \\ - x, if x < 0 \end{cases} Thus, x^{2} + 3 |x| + 2 = 0 can be written as \{\begin{cases} x^{2} + 3 x + 2 = 0, if x \geq 0 . . . (1) \\ x^{2} - 3 x + 2 = 0, if x < 0 . . . (2) \end{cases} Solving (1), we get if x \geq 0 . x^{2} + 3 x + 2 = 0 \Rightarrow x^{2} + 2 x + x + 2 = 0 \Rightarrow x (x + 2) + 1 (x + 2) = 0 \Rightarrow (x + 1) (x + 2) = 0 \Rightarrow x = - 1, - 2 But, x \geq 0 ∴ x \neq - 1, - 2 Solving (2), we get if x < 0 . x^{2} - 3 x + 2 = 0 \Rightarrow x^{2} - 2 x - x + 2 = 0 \Rightarrow x (x - 2) - 1 (x - 2) = 0 \Rightarrow (x - 1) (x - 2) = 0 \Rightarrow x = 1, 2 But, x < 0 ∴ x \neq 1, 2$

Hence, the number of real roots of the equation $x^{2} + 3 |x| + 2 = 0$ is 0.

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Question 13:

If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the roots of the quadratic equation are
______ and _______.

Answer:

Let $a x^{2} + b x + c = 0$ be a quadratic equation.

It is given that, coefficient of x² and the constant term of a quadratic equation have opposite signs
⇒ ac < 0
⇒ 4ac < 0
⇒ −4ac > 0
⇒ b²−4ac > 0
⇒ Discriminant > 0
⇒ Roots are real and distinct

Hence, If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the roots of the quadratic equation are real and distinct.

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Question 14:

If the coefficient of x² and the constant term of a quadratic equation have the same sign and if the coefficient of x term is zero, then the quadratic equation has ________ roots.

Answer:

Let $a x^{2} + 0 x + c = 0$ be a quadratic equation.

It is given that, coefficient of x² and the constant term of a quadratic equation have the same sign
⇒ ac > 0
⇒ 4ac > 0
⇒ −4ac < 0
⇒ 0²−4ac < 0
⇒ Discriminant < 0
⇒ Roots are imaginary

Hence, if the coefficient of x² and the constant term of a quadratic equation have the same sign and if the coefficient of x term is zero, then the quadratic equation has imaginary roots.

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