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Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2 + kx + 3 = 0$
(ii) $kx (x - 2) + 6 = 0$
To do:
We have to find the values of $k$ for which the given quadratic equations have equal roots.
Solution:
(i) $2x^2 + kx + 3 = 0$
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=2, b=k$ and $c=3$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=(k)^2-4(2)(3)$
$D=k^2-24$
The given quadratic equation has equal roots if $D=0$.
Therefore,
$k^2-24=0$
$k^2-(\sqrt{24})^2=0$
$(k+\sqrt{24})(k-\sqrt{24})=0$
$k+2\sqrt6=0$ or $k-2\sqrt6=0$
$k=-2\sqrt6$ or $k=2\sqrt6$
The values of $k$ are $-2\sqrt6$ and $2\sqrt6$.
(ii) $kx (x - 2) + 6 = 0$
$kx^2-2kx+6=0$
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=k, b=-2k$ and $c=6$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=(-2k)^2-4(k)(6)$
$D=4k^2-24k$
The given quadratic equation has equal roots if $D=0$.
Therefore,
$4k^2-24k=0$
$4k(k-6)=0$
$4k=0$ or $k-6=0$
$k=6$ or $k=0$ which is not possible.
The value of $k$ is $6$.