Question

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

Answer 1

I hope this answers helps you

Answer 2

Answer:

Step-by-step explanation:

(1+m²)x² + 2mcx + (c²-a²) = 0

roots of this equation are

(-2mc + √((2mc)²- 4×(1+m²)×(c²-a²)))/2(1+m²) and

(-2mc - √((2mc)²- 4×(1+m²)×(c²-a²)))/2(1+m²)

since both roots are equal therefore,

(-2mc + √((2mc)²- 4×(1+m²)×(c²-a²)))/2(1+m²)=

(-2mc - √((2mc)²- 4×(1+m²)×(c²-a²)))/2(1+m²)

denominator cancels out frm LHS and RHS

-2mc + √((2mc)²- 4×(1+m²)×(c²-a²))=-2mc - √((2mc)²- 4×(1+m²)×(c²-a²))

-2mc cancels out from LHS and RHS

√((2mc)²- 4×(1+m²)×(c²-a²))= -√((2mc)²- 4×(1+m²)×(c²-a²))

taking both terms on LHS

2√((2mc)²- 4×(1+m²)×(c²-a²)) =0

(2mc)²- 4×(1+m²)×(c²-a²)= 0

4m²c² = 4(c²-a²+m²c²-m²a²)

m²c² = c²-a²+m²c²-m²a²

m²c² cancels out from LHS and RHS, therefore

0= c²-a²-m²a²

rearranging terms

c²=a²+m²a²

c²=a²(1+m²)

hence proved