lf the expression $3 x^{2} + 2 p x y + 2 y^{2} + 2 a x - 4 y + 1$ can be resolved into two linear factors, then $p$ must be a root of the equation

x^{2} + a x + 6 = 0

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

x^{2} + 4 a x + 6 = 0

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

x^{2} + 4 a x + 2 a^{2} + 6 = 0

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

x^{2} - 4 a x + 6 = 0

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

Open in App

Solution

The correct option is D $x^{2} + 4 a x + 2 a^{2} + 6 = 0$
If $3 x^{2} + 2 p x y + 2 y^{2} + 2 a x - 4 y + 1 = 0.$ , then.
$Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$= 3 (2) (1) + 2 (- 2) (a) (p) - 3 (- 2)^{2} - 2 (a)^{2} - 1 (p)^{2} = 0$
$6 - 4 a p - 12 - 2 a^{2} - p^{2} = 0$
$p^{2} + 4 a p + 2 a^{2} + 6 = 0$
$\Rightarrow p$ is a solution of $x^{2} + 4 a x + 2 a^{2} + 6 = 0$

Suggest Corrections

Similar questions

k

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

(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0

:

x^{2} + 5 x + 6 = 0

x^{2} - 5 x + 6 = 0

x^{2} + 5 x - 6 = 0

x^{2} - 5 x - 6 = 0

(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0

x^{2} - 5 x + 6 = 0

Join BYJU'S Learning Program