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

$(2k+1)x^2 + 2(k+3)x + (k + 5) = 0$

Given:

Given quadratic equation is $(2k+1)x^2 + 2(k+3)x + (k + 5) = 0$.

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,

$a=2k+1, b=2(k+3)$ and $c=k+5$.

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

$D=[2(k+3)]^2-4(2k+1)(k+5)$

$D=4(k+3)^2-(8k+4)(k+5)$

$D=4(k^2+6k+9)-8k^2-40k-4k-20$

$D=4k^2+24k+36-8k^2-44k-20$

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

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

Therefore,

$-4k^2-20k+16=0$

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

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

$k=\frac{-5 \pm \sqrt{5^2-4(1)(-4)}}{2(1)}$

$k=\frac{-5 \pm \sqrt{25+16}}{2}$

$k=\frac{-5+\sqrt{41}}{2}$ or $k=\frac{-5-\sqrt{41}}{2}$

The values of $k$ are $k=\frac{-5+\sqrt{41}}{2}$ and $k=\frac{-5-\sqrt{41}}{2}$.   

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Updated on: 10-Oct-2022

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