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Find the values of k for which the quadratic equation $(3k + 1)x^2 + 2(k + 1)x + 1 = 0$ has equal roots. Also, find the roots.
Given:
Given quadratic equation is $(3k + 1)x^2 + 2(k + 1)x + 1 = 0$.
To do:
We have to find the value of k for which the given quadratic equation has equal roots.
Solution:
$(3k + 1)x^2 + 2(k + 1)x + 1 = 0$
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 $c=1$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=[2(k+1)]^2-4(3k+1)(1)$
$D=4(k+1)^2-4(3k+1)$
$D=4(k^2+2k+1)-12k-4$
$D=4k^2+8k+4-12k-4$
$D=4k^2-4k$
The given quadratic equation has equal roots if $D=0$.
Therefore,
$4k^2-4k=0$
$4k(k-1)=0$
$4k=0$ or $k-1=0$
$k=0$ or $k=1$
The values of k are $0$ and $1$.
For $k = 0$,
$(3k + 1)x^2 + 2(k + 1)x + 1 = 0$
$(3(0) + 1)x^2 + 2(0 + 1)x + 1 = 0$
$x^2 + 2x + 1 = 0$
$(x + 1)^2 = 0$
$x+1=0$
$x=-1$
Therefore, for $k=0$ the roots of the given quadratic equation are $1$ and $1$.
For $k = 1$,
$(3k + 1)x^2 + 2(k + 1)x + 1 = 0$
$(3(1) + 1)x^2 + 2(1 + 1)x + 1 = 0$
$4x^2 + 4x + 1 = 0$
$(2x + 1)^2 = 0$
$2x+1=0$
$2x=-1$
$x=\frac{-1}{2}$
Therefore, for $k=1$ the roots of the given quadratic equation are $\frac{-1}{2}$ and $\frac{-1}{2}$.
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