If the equation (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0 has equal roots, prove that c^2=a^2 (1+m^2 ).

Given: (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0 Here, a= (1+m^2 ), b=2 m c and c= (c^2-a^2 ) It is given that the roots of the equation are equal; therefore, we have: D=0 (b^2-4 a c )=0 (2 m c)^2-4 (1+m^2 ) (c^2-a^2 )=0 4 m^2 c^2-4 (c^2-a^2+m^2 c^2-m^2 a^2 )=0 4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2=0 -4 c^2+4 a^2+4 m^2 a^2=0 a^2+m^2 a^2=c^2 a^2 (1+m^2 )=c^2 c^2=a^2 (1+m^2 ) Hence proved.

If the equation (1+m^2 ) x^2+2 m c x+ (c^2-a^2 )=0 has equal root…

Download App

Question

If the equation $\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$ has equal roots, prove that $c^2=a^2\left(1+m^2\right)$.

✓

Answer

Given:
$\left(1+m^2\right) x^2+2 m c x+\left(c^2-a^2\right)=0$
Here,
$a=\left(1+m^2\right), b=2 m c \text { and } c=\left(c^2-a^2\right)$
It is given that the roots of the equation are equal; therefore, we have:
$D=0$
$\Rightarrow\left(b^2-4 a c\right)=0$
$\Rightarrow(2 m c)^2-4 \times\left(1+m^2\right) \times\left(c^2-a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4\left(c^2-a^2+m^2 c^2-m^2 a^2\right)=0$
$\Rightarrow 4 m^2 c^2-4 c^2+4 a^2-4 m^2 c^2+4 m^2 a^2=0$
$\Rightarrow-4 c^2+4 a^2+4 m^2 a^2=0$
$\Rightarrow a^2+m^2 a^2=c^2$
$\Rightarrow a^2\left(1+m^2\right)=c^2$
$\Rightarrow c^2=a^2\left(1+m^2\right)$
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks Question" from Quadratic Equations→Quadratic Equations — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Quadratic Equations — Maths STD 10 — Question

Answer

Need a full question paper?

Explore more

Similar questions