Question

If the equation has (1+m²)x²+2mcx+(c²-a²)=0 equal roots, prove that c²=a²(1+m²) .

Answer 1

$c^2=a^2(1+m^2)$, proved.

Step-by-step explanation:

The given quadratic equation:

$(1+m^2)x^2$ + 2mcx + ($c^2-a^2$) = 0

Here, A = $1+m^2$, B = 2mc and C = $c^2-a^2$

To prove that, $c^2=a^2(1+m^2)$.

Discriminant, D = $B^{2} -4AC$

= $(2mc)^2 - 4(1+m^2)(c^2-a^2)$

= $4m^2c^2 - 4(c^2-a^2+m^2c^2-m^2a^2)$

= $4m^2c^2 - 4c^2+4a^2-4m^2c^2+4m^2a^2$

= $- 4c^2+4a^2+4m^2a^2$

The roots are real, then

D = 0

$4c^2=4a^2+4m^2a^2$

⇒ $4c^2=4a^2(1+m^2)$

⇒ $c^2=a^2(1+m^2)$, proved.

Thus, $c^2=a^2(1+m^2)$, proved.

Answer 2

Answer:

hope this is the correct answer