Question

(a2 + b2)x2 - 2(ac + bd)x + c2+d2=0
prove that a÷b=c÷d​

Answer 1

Given:

$\fbox{\mathsf{ (a^2 + b^2)x^2 - 2(ac + bd)x + c^2+ d^2}}$

Comparing them in the form Ax² + Bx + C, we get:

A = (a² + b²)

B = -2(ac + bd)

C = c² + d²

For zeroes of the polynomial

Discriminant, D = 0

$\mathsf{\implies \: B^2 - 4AC = 0}$

$\mathsf{\implies \: [-2(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0}$

$\mathsf{\implies \: 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0}$

$\mathsf{\implies \: 4[(ac + bd)^2 - (a^2 + b^2)(c^2 + d^2)] = 0}$

$\mathsf{\implies \: 4[a^2c^2 + b^2d^2 + 2abcd - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2)] = 0}$

$\mathsf{\implies \: 4[\cancel{a^2c^2} + \cancel{b^2d^2} + 2abcd - \cancel{a^2c^2} - a^2d^2 - b^2c^2 - \cancel{b^2d^2}] = 0}$

$\mathsf{\implies \: 2abcd - a^2d^2 - b^2c^2 = 0}$

Transposing them to right side, we get

$\mathsf{\implies \: a^2d^2 + b^2c^2 - 2abcd = 0}$

$\mathsf{\implies \: (ad - bc)^2 = 0}$

Square rooting them both sides, we get

$\mathsf{\implies \: ad - bc = 0}$

$\mathsf{\implies \: ad = bc}$

$\huge\mathsf{\implies \: \frac{a}{b} = \frac{c}{d}}$

Hence Proved