Find the value of k for which the quadratic equation $x^{2} - 4 x + k = 0$ has real and equal roots.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
4

On comparing $x^{2} - 4 x + k = 0$ with standard form $a x^{2} + b x + c = 0$ , we get

$a = 1, b = - 4, c = k$

The roots of quadratic equation are real and equal only when discriminant, $D = 0$ .

$D = \sqrt{b^{2} - 4 a c} = 4^{2} - 4 k = 16 - 4 k$

$\Rightarrow 16 - 4 k = 0$

$\Rightarrow k = 4$

Suggest Corrections

Similar questions

Find the value of $k$ for which $x^{2} - 4 x + k = 0$ has coincident roots.

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

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

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

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

Join BYJU'S Learning Program