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

Answer:

An equation is of the form $a{x}^2+bx+c=0$, $a\ne 0$ is called a quadratic equation. 

(a) Given that, 
$x^2+2x+1={(4-x)}^2+3$
$\Rightarrow x^2+2x+1=16+x^2-8x+3$
$\Rightarrow 10x-18=0$, which is not a quadratic equation.

(b) Given that,
$-2x^2=(5-x)\left(2x-\frac{2}{5}\right)$
$\Rightarrow -2x^2=10x-2x^2-2+\frac{2x}{5}$
$\Rightarrow 50x+2x-10=0$
$\Rightarrow 52x-10=0$, which is not a quadratic equation.

(c) Given that,
$x^2\left(k+1\right)+\frac{3}{2}x=7$, where k = −1
$\Rightarrow x^2\left(-1+1\right)+\frac{3}{2}x=7$
$\Rightarrow 3x-14=0$, which is not a quadratic equation.

(d) Given that,
$x^3-x^2={\left(x-1\right)}^3$
$$\begin{gathered} \Rightarrow x^3-x^2=x^3-3x^2+3x-1 \\ \Rightarrow -x^2+3x^2-3x+1=0 \end{gathered}$$
$\Rightarrow 2x^2-3x+1=0$, which is a quadratic equation.

Hence, the correct answer is option (d).

Answer:

An equation is of the form $a{x}^2+bx+c=0$, $a\ne 0$ is called a quadratic equation.

(a) Given that,
$2{\left(x-1\right)}^2=4x^2-2x+1$
$$\begin{gathered} \Rightarrow 2\left(x^2+1-2x\right)=4x^2-2x+1 \\ \Rightarrow 2x^2+2-4x=4x^2-2x+1 \end{gathered}$$
$\Rightarrow 2x^2+2x-1=0$, which is a quadratic equation.

(b) Given that,
$2x-x^2=x^2+5$
$\Rightarrow 2x^2-2x+5=0$, which is a quadratic equation.

(c) Given that,
${\left(\sqrt{2}x+\sqrt{3}\right)}^2+x^2=3x^2-5x$
$\Rightarrow 2x^2+3+2\sqrt{6}x+x^2=3x^2-5x$
$\Rightarrow 5x+3+2\sqrt{6}x=0$, which is not a quadratic equation.

(d) Given that,
${\left(x^2+2x\right)}^2=x^4+3+4x^3$
$\Rightarrow x^4+4x^2+4x^3=x^4+3+4x^3$
$\Rightarrow 4x^2-3=0$, which is a quadratic equation.

Hence, the correct answer is option C.

Answer:

If α is one of the roots of the quadratic equation $a{x}^2+bx+c=0$, then x = α  satisfies the equation  the quadratic.

(a) Given, $x^2-4x+5=0$
Substituting x = 2 in $x^2-4x+5$, we get 
$2^2-4\left(2\right)+5$
$\Rightarrow 4-8+5=1\ne 0$
So we conclude, x = 2 is not a root of $x^2-4x+5=0$.

(b) Given, $x^2+3x-12=0$
Substituting x = 2 in $x^2+3x-12$, we get
${\left(2\right)}^2+3\left(2\right)-12$
$\Rightarrow 4+6-12=-2\ne 0$
So we conclude, x = 2 is not a root of $x^2+3x-12=0$.

(c) Given, $2x^2-7x+6=0$
Substituting x = 2 in $2x^2-7x+6$, we get
$2{\left(2\right)}^2-7\left(2\right)+6$
$\Rightarrow 8-14+6=14-14=0$
So we conclude, x = 2 is a root of $2x^2-7x+6=0$.

(d) Given, $3x^2-6x-2=0$
Substituting x = 2 in $3x^2-6x-2$, we get
$3{\left(2\right)}^2-6\left(2\right)-2$
$\Rightarrow 12-12-2=-2\ne 0$
So we conclude, x = 2 is not a root of $3x^2-6x-2=0$.

Hence, the correct answer is option (c).

Answer:

If α and β are the roots of the quadratic equation $a{x}^2+bx+c=0$, $a\ne 0$, then sum of roots = α + β = $\left(-1\right)\frac{\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{-b}{a}$.

(a) Given, $2x^2-3x+6=0$.
Comparing with $a{x}^2+bx+c=0$, we get
a = 2, b = −3 and c = 6.
∴ Sum of the roots = $\frac{-b}{a}=\frac{-\left(-3\right)}{2}=\frac{3}{2}$.
So, we conclude sum of the roots of given quadratic equation is not 3.

(b) Given, $-x^2+3x-3=0$.
Comparing with $a{x}^2+bx+c=0$, we get
a = −1, b = 3 and c = −3.
∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(3\right)}{-1}=3$.
So, we conclude sum of the roots of given quadratic equation is 3.

(c) Given, $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+1=0$.
Comparing with $a{x}^2+bx+c=0$, we get
$a=\sqrt{2}, b=-\frac{3}{\sqrt{2}}, c=1$

∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(-{\displaystyle \frac{3}{\sqrt{2}}}\right)}{\sqrt{2}}=\frac{3}{2}$.
So, we conclude sum of the roots of given quadratic equation is not 3.

(d) Given, $3x^2-3x+3=0$.
Comparing with $a{x}^2+bx+c=0$, we get
a = 3, b = −3 and c = 3.
∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(-3\right)}{3}=1$.
So, we conclude sum of the roots of the given quadratic equation is not 3.

Hence, the correct answer is option (b).

Answer:

Given, equation $2x^2-\sqrt{5}x+1=0.$

Now, comparing with $a{x}^2+bx+c=0$, we get

a = 2, b = $-\sqrt{5}$ and c = 1.

$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(-\sqrt{5}\right)}^2-4\times 2\times 1\\ & =& 5-8\\ & =& -3<0\end{array}$

Thus, we conclude given quadratic equation has no real roots since the discriminant is negative.

Hence, the correct answer is option (c).

Answer:

If a quadratic equation is in the form of $a{x}^2+bx+c=0$, $a\ne 0$ then
(i) If $D=b^2-4ac>0$, then its roots are distinct and real.
(ii) If $D=b^2-4ac=0$, then its roots are real and equal.
(iii) If $D=b^2-4ac<0$, then its roots are not real or imaginary roots.

(a) Given, $2x^2-3\sqrt{2}x+\frac{9}{4}=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 2, b = $-3\sqrt{2}$ and c = $\frac{9}{4}$.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(-3\sqrt{2}\right)}^2-4\times 2\times \frac{9}{4}\\ & =& 18-18\\ & =& 0\end{array}$

Thus, the equation has real and equal roots.

(b) Given $x^2+x-5=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 1, b = 1 and c = −5.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& 1^2-4\times 1\times \left(-5\right)\\ & =& 1+20\\ & =& 21>0\end{array}$

Thus, the equation has real and distinct roots.

(c) Given $x^2+3x+2\sqrt{2}=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 1, b = 3 and c = $2\sqrt{2}$
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(3\right)}^2-4\times 1\times \left(2\sqrt{2}\right)\\ & =& 9-8\sqrt{2}<0\end{array}$

Thus, equation has no real roots.

(d) Given $5x^2-3x+1=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 5, b = −3 and c = $1$
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(-3\right)}^2-4\times 5\times 1\\ & =& 9-20<0\end{array}$

Thus, the equation has no real roots.

Hence, the correct answer is option (b).

Answer:

If a quadratic equation is in the form of $a{x}^2+bx+c=0$, $a\ne 0$ then
(i) If $D=b^2-4ac>0$, then its roots are distinct and real.
(ii) If $D=b^2-4ac=0$, then its roots are real and equal.
(iii) If $D=b^2-4ac<0$, then its roots are not real or imaginary roots.

(a) Given $x^2-4x+3\sqrt{2}=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 1, b = −4 and c = $3\sqrt{2}$.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(-4\right)}^2-4\times 1\times 3\sqrt{2}\\ & =& 16-12\sqrt{2}<0\end{array}$

Thus, the equation has no real roots.

(b) Given $x^2+4x-3\sqrt{2}=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 1, b = 4 and c = $-3\sqrt{2}$.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(4\right)}^2-4\times 1\times (-3\sqrt{2})\\ & =& 16+12\sqrt{2}>0\end{array}$

Thus, the equation has real and distinct roots.

(c) Given $x^2-4x-3\sqrt{2}=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 1, b = −4 and c = $-3\sqrt{2}$.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(-4\right)}^2-4\times 1\times (-3\sqrt{2})\\ & =& 16+12\sqrt{2}>0\end{array}$

Thus, the equation has real and distinct roots.

(d) Given $3x^2+4\sqrt{3}x+4=0.$
Now, comparing with $a{x}^2+bx+c=0$, we get
a = 3, b = $4\sqrt{3}$ and c = 4.
$\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(4\sqrt{3}\right)}^2-4\times 3\times 4\\ & =& 48-48\\ & =& 0\end{array}$

Thus, the equation has real and equal roots.

Hence, the correct answer is option (a).

Answer:

Given equation, ${\left(x^2+1\right)}^2-x^2=0.$
$$\begin{gathered} \Rightarrow x^4+1+2x^2-x^2=0 \\ \Rightarrow x^4+x^2+1=0 \end{gathered}$$

Let $x^2=y$
∴ ${\left(x^2\right)}^2+x^2+1=0$
$\Rightarrow y^2+y+1=0$

Now, comparing with $a{y}^2+by+c=0$, we get.
a = 1, b = 1 and c = 1

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& {\left(1\right)}^2-4\left(1\right)\left(1\right)\\ & =& 1-4\\ & =& -3\end{array}$

Since, $D<0$.
Thus, we conclude given equation has no real roots.

Hence, the correct answer is option (c).

Answer:

Given equation, $x^2-0.4k=0$ has x = 0.2 as a root.
Now, substitute the value 0.2 in the given equation, we get
$$\begin{gathered} {\left(0.2\right)}^2-0.4k=0 \\ \Rightarrow 0.04=0.4k \\ \Rightarrow k=\frac{0.04}{0.4} \\ \Rightarrow k=0.1 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Since, $-\frac{1}{2}$ is a root of equation $x^2–kx–\frac{5}{4}=0$.
Put $x=-\frac{1}{2}$ in given equation, we get

$$\begin{gathered} \Rightarrow {\left(-\frac{1}{2}\right)}^2-k\left(-\frac{1}{2}\right)-\frac{5}{4}=0 \\ \Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0 \\ \Rightarrow \frac{1+2k-5}{4}=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2k-4=0 \\ \Rightarrow 2k=4 \\ \Rightarrow k=2 \end{gathered}$$

Thus, the value of k = 2.

Hence, the correct answer is option (b).

Answer:

An equation is of the form $a{x}^2+bx+c=0$, $a\ne 0$ is called a quadratic equation.

(a) Given that,
3(x + 1)² = 2x² + x + 4
$$\begin{gathered} \Rightarrow 3\left(x^2+1+2x\right)=2x^2+x+4 \\ \Rightarrow 3x^2+3+6x=2x^2+x+4 \\ \Rightarrow x^2+5x-1=0 \end{gathered}$$
which is a quadratic equation.

(b) Given that,
5x + 2x² = x² + 9
$\Rightarrow x^2+5x-9=0$, which is a quadratic equation.

(c) Given that,
${\left(x^2-2x\right)}^2=x^4+3+4x^2$
$\Rightarrow x^4+4x^2-4x^3=x^4+3+4x^2$
$\Rightarrow 4x^3+3=0$, which is not a quadratic equation.

(d) Given that,
${\left(\sqrt{2}x+\sqrt{3}\right)}^2=2x^2–3x$
$$\begin{gathered} \Rightarrow 2x^2+3+2\sqrt{6}x=2x^2-3x \\ \Rightarrow \left(2\sqrt{6}+3\right)x+3=0 \end{gathered}$$
, which is not a quadratic equation.

Disclaimer: Both (c) and (d) options are correct.

Answer:

If α is one of the root of quadratic equation $a{x}^2+bx+c=0$, then x = α satisfies the equation $a{\alpha }^2+b\alpha +c=0.$

(a) Given, x² – 4x + 3 = 0
Substituting x = 3 in $x^2-4x+3$, we get 
$$\begin{gathered} 3^2-4\left(3\right)+3 \\ =9-12+3 \\ =0 \end{gathered}$$
So we conclude, x = 3 is a root of x² – 4x + 3 = 0.

(b) Given, x² + 4x + 3 = 0
Substituting x = 3 in x² + 4x + 3, we get
$$\begin{gathered} 3^2+4\left(3\right)+3 \\ =9+12+3 \\ =24\ne 0 \end{gathered}$$
So we conclude, x = 3 is not a root of x² + 4x + 3 = 0.

(c) Given, x² + 5x + 6 = 0
Substituting x = 3 in x² + 5x + 6, we get
$$\begin{gathered} {\left(3\right)}^2+5\left(3\right)+6 \\ =9+15+6 \\ =30\ne 0 \end{gathered}$$
So we conclude, x = 3 is not a root of x² + 5x + 6 = 0.

(d) Given, x² + 7x + 12 = 0
Substituting x = 3 in x² + 7x + 12, we get
$$\begin{gathered} {\left(3\right)}^2+7\left(3\right)+12 \\ =9+21+12 \\ =42 \end{gathered}$$
So we conclude, x = 3 is not a root of x² + 7x + 12 = 0.

Hence, the correct answer is option (a).

Answer:

A quadratic equation can have atmost two roots, zero, one and two.

Hence, the correct answer is option (b).

Answer:

The quadratic equation (x + 2)² = 0 can be written as $x^2+4x+4=0$.

On comparing with $a{x}^2+bx+c=0$, we get

$a=1, b=4, c=4$

$\begin{array}{rcl}\mathrm{Discriminant}, D& =& b^2-4ac\\ & =& 4^2-4\times 1\times 4\\ & =& 16-16\\ & =& 0\end{array}$

Hence, the correct answer is option (d).

Answer:

The given quadratic equation  $16x^2+4kx+9=0$, has equal roots.

Here, $a=16, b=4k \mathrm{and} c=9$.

As we know that $D=b^2-4ac$

Putting the values of $a=16, b=4k \mathrm{and} c=9$.

$$\begin{gathered} D={\left(4k\right)}^2-4\left(16\right)\left(9\right) \\ =16k^2-576 \end{gathered}$$

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

Thus, $16k^2-576=0$
$$\begin{gathered} \Rightarrow k^2-36=0 \\ \Rightarrow (k+6)(k-6)=0 \\ \Rightarrow k+6=0 \mathrm{or} k-6=0 \\ \Rightarrow k=-6 \mathrm{or} k=6 \end{gathered}$$

Therefore, the value of k is 6, −6.

Hence, the correct option is (c).

Answer:

Since, y = 1 is a root of the equations $a{y}^2+ay+3=0$.
So, it satisfies the given equation.

$$\begin{gathered} \therefore a{\left(1\right)}^2+a\left(1\right)+3=0 \\ \Rightarrow 2a+3=0 \\ \Rightarrow a=-\frac{3}{2} ...\left(1\right) \end{gathered}$$

Since, y = 1 is a root of the equations $y^2+y+b=0$.
So, it satisfies the given equation.

$$\begin{gathered} \therefore {\left(1\right)}^2+\left(1\right)+b=0 \\ \Rightarrow 2+b=0 \\ \Rightarrow b=-2 ...\left(2\right) \end{gathered}$$

From (1) and (2),
$$\begin{gathered} ab=\left(-\frac{3}{2}\right)\left(-2\right) \\ =3 \end{gathered}$$

Thus, ab is equal to 3.

Hence, the correct option is (a).

Answer:

Let be the roots of quadratic equation in such a way that

Here,

Then , according to question sum of the roots

And the product of the roots

Therefore, value of other root be

Thus, the correct answer is

Answer:

Let be the roots of quadratic equation in such a way that

Here,

Then , according to question sum of the roots

And the product of the roots

Putting the value of in above

Putting the value of k in

Therefore, value of other root be

Thus, the correct answer is

Answer:

Let be the roots of quadratic equation in such a way that

Then, according to question sum of the roots

And the product of the roots

As we know that the quadratic equation

Putting the value of in above

Therefore, the require equation be

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal

Then find the value of c.

Let be two roots of given equation

And,

Then, as we know that sum of the roots

And the product of the roots

According to question, sum of the roots = product of the roots

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are zero.

Then find the value of .

Here,

As we know that

Putting the value of

The given equation will have zero roots, if

Therefore, the value of

Thus, the correct answer is

Answer:

is the common roots given quadric equation are , and

Then find the value of q.

Here, ….. (1)

….. (2)

Putting the value of in equation (1) we get

Now, putting the value of in equation (2) we get

Then,

Thus, the correct answer is

Answer:

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

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal

Then find the value of c.

Let be two roots of given equation

Then, as we know that sum of the roots

And the product of the roots

Putting the value of

Therefore, the value of

Thus, the correct answer is

Answer:

Let

Squaring both sides we get

The value of x cannot be negative.

Thus, the value of x = 3

Therefore, the correct answer is

Answer:

2 is the common roots given quadric equation are , and

Then find the value of q.

Here, ….. (1)

….. (2)

Putting the value of in equation (1) we get

Now, putting the value of in equation (2) we get

Then,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

Answer:

Given that p and q be the roots of the equation

Then find the value of p and q.

Here,

p and q be the roots of the given equation

Therefore, sum of the roots

….. (1)

Product of the roots

As we know that

Putting the value of in equation (1)

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal

Then find the value of c.

Let be two roots of given equation

Then, as we know that sum of the roots

And the product of the roots

Putting the value of

Therefore, the value of

Thus, the correct answer is

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 real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

Answer:

Let be the roots of quadratic equation in such a way that

Here,

Then,

according to question sum of the roots

….. (1)

And the product of the roots

….. (2)

Putting the value of in equation (2)

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal

Then find the value of k.

Let be two roots of given equation

And,

Then, as we know that sum of the roots

And the product of the roots

According to question, sum of the roots product of the roots

Therefore, the value of

Thus, the correct answer is

Answer:

Let be the roots of quadratic equation in such a way that

Here,

Then , according to question sum of the roots

And the product of the roots

Therefore, value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are distinct.

Then find the value of a.

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal.

Then find roots of given equation.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

So, putting the value of k in quadratic equation

When then equation be and when then

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are distinct.

Then find the value of a.

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation are , and roots are real.

Then find the value of a.

Here, ….. (1)

….. (2)

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, putting the value of in equation (2) we get

The value of satisfying to both equations

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

Answer:

The given quadric equation is , and does not have real roots.

Then find the value of b.

Here,

As we know that

Putting the value of

The given equation does not have real roots, if

Therefore, the value of

Thus, the correct answer is

Answer:

Given that the equation _.

For given equation to have real roots, discriminant (D) ≥ 0

⇒ b² − 4a ≥ 0

⇒ b² ≥ 4a

⇒ b ≥ 2√a

Now, it is given that a and b can take the values of 1, 2, 3 and 4.

The above condition b ≥ 2√a can be satisfied when

i) b = 4 and a = 1, 2, 3, 4

ii) b = 3 and a = 1, 2

iii) b = 2 and a = 1

So, there will be a maximum of 7 equations for the values of (a, b) = (1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3) and (1, 2).

Thus, the correct option is (b).

Answer:

As we know that the number of quadratic equations having real roots and which do not change by squaring their roots is 2.

Thus, the correct answer is

Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have no real roots, if

Thus, the correct answer is

Answer:

The given quadric equation is , and are roots of given equation.

And,

Then, as we know that sum of the roots

…. (1)

And the product of the roots

…. (2)

Squaring both sides of equation (1) we get

Putting the value of , we get

Putting the value of , we get

Therefore, the value of

Thus, the correct answer is

Answer:

The given quadric equation is , and their roots are a and b

Then find the value of

Let be two roots of given equation

And,

Then, as we know that sum of the roots

And the product of the roots

Putting the value of a in above

Therefore, the value of

Thus, the correct answer is

Answer:

(i) When no barrel is produced, x = 0

P(0) = −10(0)² + 3500(0) − 66000

⇒ P(0) = − 66000

Thus, when no barrel is produced, the company will bear a loss of â‚¹66,000.

Hence, the correct answer is option (c).

(ii) The break-even point is the No-profit No-loss condition, i.e., at the break-even point, the company's profit is zero.

∵ P(x) = 0

⇒ −10x² + 3500x − 66000 = 0

⇒ x² − 350x + 6600 = 0

​⇒ x² − 330x − 20x + 6600 = 0

​⇒ x(x − 330) − 20(x − 330) = 0

⇒ (x − 20)(x − 330) = 0

⇒ x = 20, 330

The maximum oil produced by company is 300 barrels. So, x = 330 will be rejected.

Thus, the break-even point is x = 20

Hence, the correct answer is option (b).

(iii) On producing 100 barrels, the company, x = 100

P(100) = −10(100)² + 3500(100) − 66000

⇒ P(100) = −100000 + 350000 − 66000

⇒ P(100) = 184000

Thus, on producing 100 barrels, the company earns profit of ₹184,000.

Hence, the correct answer is option (b).

(iv)  If company produces 400 barrels, then x = 400

P(400) = −10(400)² + 3500(400) − 66000

⇒ P(400) = −1600000 + 1400000 − 66000

⇒ P(400) = −266000

Thus, on producing 400 barrels, the company earns loss of ₹266,000.

Hence, the correct answer is option (b).

(v) The profit function is a quadratic equation.

Thus, the graph of the profit function is a parabola.

Hence, the correct answer is option (c).

 

Answer:

(i) The length of the rice land is x metres.

The length of wheatland is 3 metres more than twice the length of riceland.

The length of the wheatland is 2x + 3.



Thus, the length of the field = x + 2x + 3 = (3x + 3) m

Hence. the correct answer is option (b).

(ii) The figure given below shows the measurements of each side of the field.



The perimeter of the field = 4x + 2(2x + 3) = (8x + 6) m

Hence, the correct answer is option (a).

(iii) The total area of the field = 1260 m²

⇒ x² + x(2x + 3) = 1260

⇒ x² + 2x² + 3x = 1260

⇒ 3x² + 3x​ − 1260 = 0

⇒ x² + x​ − 420 = 0

⇒ x² + 21x​ − 20x − 420 = 0

⇒ (x + 21)(x​ − 20) = 0

⇒ x = −21​, 20

As x is the length then it can not be negative.

Thus, x = 20

Hence, the correct answer is option (c)

(iv) The area of the wheatland = x(2x + 3)

= 2x² + 3x

= 2(20)² + 3(20)

= 860 m²

Hence, the correct answer is option (d).

(v) The area of riceland = x² = 400

The ratio of the areas of the wheat and rice land = 860 : 400 = 43 : 20

Thus, the ratio of the areas of the wheat and rice land is 43 : 20.

Hence, the correct answer is option (a).

Answer:

If one root of the quadratic equation is irrational, then another root is always irrational when the coefficients are real and the other root is conjugate of the first root.
Thus, statement-2 is true.

 If $2+\sqrt{3}$ is a root of a quadratic equation with rational coefficients, then its other root is $2-\sqrt{3}.$
The roots of a quadratic equation ax² + bx + c = 0 are given by $\frac{-b\pm \sqrt{D}}{2} , \mathrm{where} D=b^2-4ac \mathrm{and} D\ge 0$.
So, when b = −2 and D = 3 for a quadratic equation then its roots will be $2+\sqrt{3}$ and $2-\sqrt{3}.$
Thus, statement-1 is true.

Thus, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

The sum of two real numbers is positive, then both numbers need not be positive.
For example: (−3) + 8 = 5
Thus, statement-2 is false.

We have,
x² + px + q = 0        .....(1)

​x² + rx + s = 0         .....(2)

Let D₁ and D₂ be the discriminants of equations (1) and (2), then

D₁ = p² − 4q

D₂ = r² − 4s
 

Answer:

We have, ax² + bx + c = 0

So, discriminant D = b² − 4ac

Given: a + b + c = 0

⇒ b = −(a + c)

∴ D = [−(a + c)]² − 4ac

⇒ D = a² + c² + 2ac − 4ac

⇒ D = a² + c² − ac

⇒ D = (a − c)² ≥ 0

Thus, if a + b + c = 0, then ax² + bx + c = 0 has real roots.
Thus, statement-1 is true.

It is also true that if one root of a quadratic equation is real, then the other root is also real.
​Thus, statement-2 is true.

Thus, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

We have, ax² + bx + c = 0

So, discriminant D = b² − 4ac

Given: a – b + c = 0

⇒ b = (a + c)

∴ D = (a + c)² − 4ac

⇒ D = a² + c² + 2ac − 4ac

⇒ D = a² + c² − 2ac

⇒ D = (a − c)² ≥ 0

Thus, if a – b + c = 0, then ax² + bx + c = 0 has real roots.

So, Statement-1 is true.

We have, x² – x + 1 = 0

So, discriminant D = (−1)² − 4(1)(1)

⇒ D = 1 − 4 = −3 < 0

Thus, roots of x² – x + 1 = 0 are not real.

So, Statement-2 is true.

Thus, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).