Question

Of the five quadratics listed below, four of them have two distinct roots. The fifth quadratic has a repeated root. Find the value of the repeated root.

Answer 1

Answer:

The value of the repeated root. is x=0.60

see the explanation

Step-by-step explanation:

we know that

In a quadratic equation of the form

$ax^{2} +bx+c=0$

The discriminant is equal to

$D=b^2-4ac$

If D=0 ----> The quadratic equation has repeated root

If D>0 ----> The quadratic equation has two different real  solutions

If D<0 ----> The quadratic equation has two different complex  solutions

Verify each case

step 1

we have

$-x^2+18x+81$

To find the roots equate the equation to zero

$-x^2+18x+81=0$

so

$a=-1\\b=18\\c=81$

Find the discriminant D

$D=18^2-4(-1)(18)=396$

$D>0$

therefore

The quadratic equation has two distinct roots

step 2

we have

$3x^2-3x-168$

To find the roots equate the equation to zero

$3x^2-3x-168=0$

so

$a=3\\b=-3\\c=-168$

Find the discriminant D

$D=-3^2-4(3)(-168)=2,025$

$D>0$

therefore

The quadratic equation has two distinct roots

step 3

we have

$x^2-4x-4$

To find the roots equate the equation to zero

$x^2-4x-4=0$

so

$a=13\\b=-4\\c=-4$

Find the discriminant D

$D=-4^2-4(13)(-4)=224$

$D>0$

therefore

The quadratic equation has two distinct roots

step 4

we have

$25x^2-30x+9$

To find the roots equate the equation to zero

$25x^2-30x+9=0$

so

$a=25\\b=-30\\c=9$

Find the discriminant D

$D=-30^2-4(25)(9)=0$

$D=0$

therefore

The quadratic equation has repeated roots

step 5

Find the value of the repeated root

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

we have

$a=25\\b=-30\\c=9\\D=0$

substitute

$x=\frac{-(-30)\pm\sqrt{0}} {2(25)}$

$x=\frac{30\pm0} {50}$

$x=\frac{30} {50}$

$x=0.60$