If the equation ${p x}^{2} + 2 x + p = 0$ has two distinct roots if.

p = \pm 1

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

p = 0

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

p = 0, 1

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

p = - 1, 0

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

Open in App

Solution

The correct option is B $p = \pm 1$
$Δ \geq 0$
$b^{2} - 4 a c \geq 0$
$a = p, b = 2, c = p$
$4 - 4 p^{2} \geq 0, 1 - p^{2} \geq 0, 1 \geq p^{2}$
$p = \pm 1$

Suggest Corrections

Similar questions

If the equation px² – 6x – 2 = 0 has real roots then, ________

x^{2} - 3 x + p = 0

p x^{2} - 2 \sqrt{5} p x + 15 = 0

p

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

x^{2} + 8 x + p = 0

Join BYJU'S Learning Program