If the discriminate of $x^{2} - 10 x + (2 k - 1) = 0$ is $40$ then the value of $k$ is

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Solution

On comparing given equation with $a x^{2} + b x + c = 0$ we get,
$a = 1, b = - 10$ and $c = 2 k - 1$

It is given that discriminate is equal to $40.$ So,
$\begin{matrix} D & = b^{2} - 4 a c 40 & = {(- 10)}^{2} - 4 (1) (2 k - 1) 40 & = 100 - 4 (2 k - 1) 4 (2 k - 1) & = 100 - 40 8 k - 4 & = 60 8 k & = 64 k & = 8 \end{matrix}$

So, value of $k$ is equal to $8.$

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