Show that the roots of the equation $3 p^{2} x^{2} - 2 p q x + q^{2} = 0$ are not real.

Open in App

Solution

For non real roots
$b^{2} - 4 a c < 0$
$b = - 2 p q$ $a = 3 p^{2}$ $c = q^{2}$
$4 p^{2} q^{2} - 4 (3 p^{2}) q^{2}$
$= - 8 p^{2} q^{2} < 0$
Since $p^{2} q^{2} < 0$
Therefore roots are unreal

Suggest Corrections

Similar questions

2 a^{2} x^{2} - 2 a b x + b^{2} = 0

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

If a,b,p,q are non-zero real numbers, then the two equations $2 a^{2} x^{2} - 2 a b x + b^{2} = 0$ and $p^{2} x^{2} + 2 p q x + q^{2} = 0$ have

x^{2} - 2 p x + p^{2} - q^{2} + 2 q r - r^{2} = 0

{a x}^{2} + 2 b x + c = 0

a, b, c

m

n

m^{2} > n > 0

{a x}^{2} + 2 m b x + n c = 0

x^{2} - 2 p x + p^{2} - q^{2} - r^{2} = 0

Join BYJU'S Learning Program