Question

(1+m)x²+(2mc)x+(c²-a²)=0 prove that c²-a²(1+m)​

Answer 1

Answer:

hey dear

here is your answer

Answer 2

Step-by-step explanation:

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

comparing with ax^2 + bx + c = 0

a = ( 1 + m ^2) , b = 2mc , c = ( c^2 - a^2 )

roots are equal

b^2- 4ac = 0

( 2mc )^2 - 4 [ ( 1 + m^2 ) ( c^2 - a^2 ) = 0

4m^2c^2 - 4 [ c^2 - a^2 + m^2c^2 - m^2a^2 ] = 0

4m^2c^2 -4c^2 +4 a^2 - 4m^2c^2 + 4m^2a^2 = 0

- 4c^2 - 4a^2 + 4m^2a^2 = 0

- 4 [ c^2 + a^2 - m^2 a^2 ] = 0

c^2 + a^2 - m^2 a^2 = 0

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