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

$(k+1)x^2 + 2(k+3)x + (k+8) = 0$

Given:

Given quadratic equation is $(k+1)x^2 + 2(k+3)x + (k+8) = 0$.

To do:

We have to find the values of k for which the roots are real and equal.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=k+1, b=2(k+3)$ and $c=k+8$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

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

$D=4(k+3)^2-(4k+4)(k+8)$

$D=4(k^2+6k+9)-4k^2-32k-4k-32$

$D=4k^2+24k+36-4k^2-36k-32$

$D=-12k+4$

The given quadratic equation has real and equal roots if $D=0$.

Therefore,

$-12k+4=0$

$12k=4$

$k=\frac{4}{12}$

$k=\frac{1}{3}$

The value of $k$ is $\frac{1}{3}$. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

35 Views

Kickstart Your Career

Get certified by completing the course

Get Started