Question

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

The discriminant of the equation should be zero since there are two equal roots.

Now,

D=4m²c²-4(1+m²)(c²-a²)

=4m²c²-4(c²-a²+m²c²-m²a²)=0

Divide each side by four:

→ m²c²-c²+a²-m²c²+m²a²=0

→ a²-c²+m²a²=0

Move side of c²

→ c²=a²(1+m²)

Answer 2

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

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.