If the roots of $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$ are equal then $a, b, c$ are in

A.P.

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

G.P.

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

H.P.

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

None of these

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

Open in App

Solution

The correct option is B G.P.
Given eqn

$(a^{2} + b^{2}) x^{2} - (2 a b + 2 b c) x + b^{2} + c^{2} = 0$

$\Rightarrow (a x - b)^{2} + (b x - c)^{2} = 0$

This is possible only when $a x = b$ and $b x = c$

$x = \frac{b}{a} = \frac{c}{b}$

$b^{2} = a c$

So $a, b, c$ are in G.P.

Suggest Corrections

Similar questions

a, b, c, x

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

a, b, c

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

b = \frac{2 a c}{a + c}

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

b = \frac{1}{2} (a + c)

b = \frac{2 a c}{(a + c)}

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) \geq 0

x ϵ R

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

Join BYJU'S Learning Program