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

$4x^2 – 2(k + 1)x + (k + 4) = 0$

Given:

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

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,

$a=4, b=-2(k+1)$ and $c=k+4$.

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

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

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

$D=4(k^2+2k+1)-16k-64$

$D=4k^2+8k+4-16k-64$

$D=4k^2-8k-60$

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

Therefore,

$4k^2-8k-60=0$

$4(k^2-2k-15)=0$

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

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

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

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

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

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

The values of $k$ are $-3$ and $5$.  

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Updated on: 10-Oct-2022

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