Question

If the following quadratic equation has two equal and real roots then find the value of k :
$16x^2-40x+\frac{k-1}{2}=0$

Answer 1

$16x^2-40x+\frac{k-1}{2}=0$

For equal & real roots, $D=0$

$D=(40{)}^2-4(16\left)\right(\frac{k-1}{2})=0$

$\Rightarrow \frac{4(k-1)}{2}=\frac{(40{)}^2}{16}$

$\Rightarrow k-1=50$

$\Rightarrow k=51$