Find the values of $k$ for which the given equation has real and equal roots
$2 x^{2} - k x + 1 = 0$

Open in App

Solution

The given equation is $2 x^{2} - k x + 1 = 0$

Here, $a = 2, b = - k, c = 1$

$∴ D = b^{2} - 4 a c = {(- k)}^{2} - 4 \times 2 \times 1 = k^{2} - 8$

The given equation will have real and equal roots, if

$D = 0 \Rightarrow k^{2} - 8 = 0 \Rightarrow k = \pm 2 \sqrt{2}$

Suggest Corrections

Similar questions

2 x^{2} + k x + 3 = 0

2 x^{2} + k x + 2 = 0

\leq

2 x^{2} + k x + 4 = 0

k

\geq 4 \sqrt{2}

k \leq - 4 \sqrt{2}

k

(i) k x^{2} + 2 x + 1 = 0

Join BYJU'S Learning Program