a) write a second degree equation with a negative constant term
b) determine the nuber of solutions of the above equation
c) write another equation which has exactly one solution
Answer :
a ) Second-degree equations involve at least one variable that is squared, or raised to a power of two. One of the most well-known second-degree equations is the quadratic ax2 + bx + c = 0 where a, b, and c are constants and a is not equal 0.
So, we can write a second degree equation witha negative constant term ,
x2 + 5x - 6 = 0 ( Ans )
b ) We know number of solution depends of the value of b2 - 4ac
Condition I ) b2 - 4ac > 0 , then our equation gives two real solution .
Condition II ) b2 - 4ac < 0 , then our equation gives two Complex solution .
Condition III ) b2 - 4ac = 0 , then our equation gives one solution .
Here we have equation x2 + 5x - 6 = 0
So,
a = 1 , b = 5 and c = - 6
So,
b2 - 4ac = 52 - 4 ( 1 ) ( - 6 ) = 25 + 24 = 49 So,
b2 - 4ac > 0 , then our equation gives two real solution . ( Ans )
We can easily find out its solution by using splitting the middle term method , As :
x 2 + 6x - x - 6 = 0
x ( x + 6 ) - 1 ( x + 6 ) = 0
( x - 1 ) ( x + 6 ) = 0
So,
x = 1 , -6 ( These are the two real solution for our equation )
c ) That's when the two solutions happen to be the same number.
Let our equation gives two solution As x = 3 and x = 3
So
Our equation will be
( x - 3 ) ( x - 3 ) = 0
x2 - 6x + 9 = 0 ( Ans )
a ) Second-degree equations involve at least one variable that is squared, or raised to a power of two. One of the most well-known second-degree equations is the quadratic ax2 + bx + c = 0 where a, b, and c are constants and a is not equal 0.
So, we can write a second degree equation witha negative constant term ,
x2 + 5x - 6 = 0 ( Ans )
b ) We know number of solution depends of the value of b2 - 4ac
Condition I ) b2 - 4ac > 0 , then our equation gives two real solution .
Condition II ) b2 - 4ac < 0 , then our equation gives two Complex solution .
Condition III ) b2 - 4ac = 0 , then our equation gives one solution .
Here we have equation x2 + 5x - 6 = 0
So,
a = 1 , b = 5 and c = - 6
So,
b2 - 4ac = 52 - 4 ( 1 ) ( - 6 ) = 25 + 24 = 49 So,
b2 - 4ac > 0 , then our equation gives two real solution . ( Ans )
We can easily find out its solution by using splitting the middle term method , As :
x 2 + 6x - x - 6 = 0
x ( x + 6 ) - 1 ( x + 6 ) = 0
( x - 1 ) ( x + 6 ) = 0
So,
x = 1 , -6 ( These are the two real solution for our equation )
c ) That's when the two solutions happen to be the same number.
Let our equation gives two solution As x = 3 and x = 3
So
Our equation will be
( x - 3 ) ( x - 3 ) = 0
x2 - 6x + 9 = 0 ( Ans )