Question

If the equation [1+m^2]x^2+2mcx+[c^2-a^2]=0 has equal roots, prove that c^2=a^2[1+m^2]

Answer 1

Here is the solution to your question-

Given equation is:

(1 + m 2) x 2 + 2mcx + (c 2 – a 2) = 0

To prove: c 2 = a 2 (1 + m 2)

Proof: It is being given that equation has equal roots, therefore

D = b 2 – 4ac = 0 ... (1)

From the above equation, we have

a = (1 + m 2)

b = 2mc

and c= (c 2 – a 2)

Putting values of a, b and c in (1), we get

D = (2mc)2 – 4 (1 + m 2) (c 2 – a 2) = 0

⇒ 4m 2 c 2 – 4 (c 2 + c 2 m 2 – a 2 – a 2 m 2) = 0

⇒ 4m 2 c 2 – 4c 2 – 4c 2 m 2 + 4a 2 + 4a 2 m 2 = 0

⇒ – 4c 2 + 4a 2 + 4a 2 m 2 = 0

⇒ 4c 2 = 4a 2 + 4a 2 m 2

⇒ c 2 = a 2 + a 2 m 2

⇒ c 2 = a 2 (1 + m 2)

Hence proved.

Answer 2

[ 1 + m^2 ] x^2 + 2mcx + [ c^2 - a^2 ]

here

a = 1 + m^2

b = 2mc

c = c^2 - a^2

= ( c - a )( c + a )

since the equation has equal roots

b^2 = 4ac

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

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

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

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

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

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

4c^2 = 4a^2 ( 1 - m^2 )

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