Find the values of k for which the following equations have real and equal roots:
$(k+1)x^2 - 2(k - 1)x + 1 = 0$

Given:

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

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,

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

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

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

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

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

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

$D=4k^2-12k$

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

Therefore,

$4k^2-12k=0$

$4(k^2-3k)=0$

$k^2-3k=0$

$k(k-3)=0$

$k=0$ or $k=3$

The values of $k$ are $0$ and $3$.  

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Updated on: 10-Oct-2022

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