Question

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

Answer 1

Solution

Given :-

• Equation, (1+m²)x² + 2mc x + (c² - a²) = 0
• Roots are equal

Show That:-

• c² = a²(1 + m²)

Explanation

Let,

• Roots are p & q

Then, according to question,

• p = q

Using Formula

$\dag\boxed{\underline{\tt{\orange{\:Sum\:of\: Roots\:=\:\dfrac{-(Coefficient\:of\:x)}{(Coefficient\:of\:x^2)}}}}}$

$\dag\boxed{\underline{\tt{\red{\: product\:of\:roots\:=\:\dfrac{(constant\:part)}{(Coefficient\:of\:x^2)}}}}}$

So, Now

==> Product of roots = (c² - a²)/(1 + m²)

==> p.q = (c² - a²)/(1 + m²)

We Know,

• p = q

==> p.p = (c² - a²)/(1 + m²)

==> p² = (c² - a²)/(1 + m²) ___________(1)

Again,

==> Sum of roots = -2mc/(1 + m²)

==> p + q = -2mc/(1 + m²)

We know,

• p = q

==> p + p = - 2mc/(1 + m²)

==> 2p = - 2mc/(1 + m²)

Or,

==> p = -mc/(1 + m²)

Keep Value of p in equ(1)

==> [ -mc/(1 + m²) ]² = (c² - a²)/(1 + m²)

==> m²c²/(c² - a²) = (1 + m²)²/(1 + m²)

==> m²c²/(c² - a²) = (1 + m²)

==> m²c² = (c² - a²)(1 + m²)

==> m²c² = c² + m²c² - a² - a²m²

==> a²(1 + m²) = c²

Or,

==> c² = a²(1 + m²)

That's proved.

__________________

Answer 2

Hey there,

Conditions implemented → For equal roots

Hence, D=0 , b² = 4ac

proceeding in that manner,

Given equation,

(1 + m²)x² + 2mcx + ( c² - a² ) =0

For equal roots,

4m²c² = 4( 1 + m² ) ( c² - a² )

4m²c² = 4c² - 4a² + 4c²m² - 4a²m²

4c² - 4a²m² - 4a² = 0

4 [ c² - a²m² - a² ] = 0

c² = a²m² + a²

c² = a² [ 1 + m² ]  

Hope this helps!