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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 $6$.

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