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

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

Given:

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

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,

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

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

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

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

$D=4(k^2+2k+1)-12k^2-4k$

$D=4k^2+8k+4-12k^2-4k$

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

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

Therefore,

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

$-4(2k^2-k-1)=0$

$2k^2-k-1=0$

$2k^2-2k+k-1=0$

$2k(k-1)+1(k-1)=0$

$(2k+1)(k-1)=0$

$2k+1=0$ or $k-1=0$

$2k=-1$ or $k=1$

$k=\frac{-1}{2}$ or $k=1$

The values of $k$ are $\frac{-1}{2}$ and $1$. 

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Updated on: 10-Oct-2022

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