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