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

$kx^2 + kx + 1 = -4x^2-x$

Given:

Given quadratic equation is $kx^2 + kx + 1 = -4x^2-x$.

To do:

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

Solution:

$kx^2 + kx + 1 = -4x^2-x$

$kx^2 + kx + 1 + 4x^2 + x = 0$

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

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

$a=k+4, b=(k+1)$ and $c=1$.

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

$D=(k+1)^2-4(k+4)(1)$

$D=(k^2+2k+1)-4k-16$

$D=k^2-2k-15$

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

Therefore,

$k^2-2k-15=0$

$k^2-5k+3k-15=0$

$k(k-5)+3(k-5)=0$

$(k+3)(k-5)=0$

$k+3=0$ or $k-5=0$

$k=-3$ or $k=5$

The values of $k$ are $-3$ and $5$. 

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Updated on: 10-Oct-2022

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