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

$5x^2 - 4x + 2+ k(4x^2 - 2x -
= 0$

Given:

Given quadratic equation is $5x^2 - 4x + 2+ k(4x^2 - 2x - 1) = 0$.

To do:

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

Solution:

$5x^2 - 4x + 2+ k(4x^2 - 2x - 1) = 0$

$5x^2-4x+2+4kx^2-2kx-k=0$

$(5+4k)x^2+(-4-2k)x+2-k=0$

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

$a=4k+5, b=-(2k+4)$ and $c=2-k$.

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

$D=[-(2k+4)]^2-4(4k+5)(2-k)$

$D=(2k+4)^2-(16k+20)(2-k)$

$D=(4k^2+16k+16)-32k+16k^2-40+20k$

$D=4k^2+16k+16+16k^2-12k-40$

$D=20k^2+4k-24$

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

Therefore,

$20k^2+4k-24=0$

$4(5k^2+k-6)=0$

$5k^2+k-6=0$

$5k^2-5k+6k-6=0$

$5k(k-1)+6(k-1)=0$

$(k-1)(5k+6)=0$

$k-1=0$ or $5k+6=0$

$k=1$ or $5k=-6$

$k=1$ or $k=\frac{-6}{5}$

The values of $k$ are $\frac{-6}{5}$ and $1$.  

Tutorialspoint

Updated on: 10-Oct-2022

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