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# Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.

$kx (x - 2) + 6 = 0$

Given:

Given quadratic equation is $kx (x - 2) + 6 = 0$

To do:

We have to find the value of k for which the given quadratic equation has equal roots.

Solution:

$kx (x - 2) + 6 = 0$

$kx^2-2kx+6=0$

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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