For what value of $k$ , are the roots of the quadratic equation, $k x (x - 2) + 6 = 0$ equal?

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Solution

Value of $k = (?)$ , equation $k x (x - 2) + 6 = 0$ has equal roots.
$⟹ k x (x - 2) + 6 = 0$
$k x^{2} - 2 k x + 6 = 0$
$⟹$ two roots of quadratic equation
$α, β = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$α, β = \frac{+ 2 k \pm \sqrt{(- 2 k)^{2} - 4 \times 6 \times k}}{2 \times k}$
$α, β = 2 k \frac{\pm \sqrt{4 k^{2} - 24 k}}{2 k}$
$⟹$ roots are equal, $α = β$
$⟹$ $\frac{2 k + \sqrt{4 k^{2} - 24 k}}{2 k} = \frac{2 k - \sqrt{4 k^{2} - 24 k}}{2 k}$
$⟹ 2 \sqrt{4 k^{2} - 24 k} = 0$
$⟹ 4 k^{2} - 24 k = 0$
$⟹$ $4 k (k - 6) = 0$
$⟹ k = 0$ or $k = 6$
$⟹ k = 6$ $[∵ k \neq 0]$ as $k = 0$ doesn't satisfy the equation

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