#### Question: If the roots of the equation (a^{2} + b^{2}) x^{2} – 2(ac + bd) x + (c^{2} + d^{2}) = 0 are equal, prove that \\frac{a}{b}) = \\frac{c}{d}).

#### Video Explanation

#### Explanatory Answer

Roots of (a^{2} + b^{2})x^{2} – 2(ac + bd)x + c^{2} + d^{2} = 0 are equal.

For a quadratic equation of the form Px^{2} + Qx + R = 0, roots are real and equal when Q^{2} – 4PR = 0.

In this equation P = (a^{2} + b^{2}), Q = -2(ac + bd), and R = c^{2} + d^{2}

Therefore, (-2(ac + bd))^{2} - 4(a^{2} + b^{2})( c^{2} + d^{2}) = 0

i.e., 4(a^{2}c^{2} + b^{2}d^{2} + 2abcd) – 4(a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}) = 0

= 4a^{2}c^{2} + 4b^{2}d^{2} + 8abcd - 4a^{2}c^{2} - 4a^{2}d^{2} - 4b^{2}c^{2} - 4b^{2}d^{2} = 0

4a^{2}d^{2} + 4b^{2}c^{2} - 8abcd = 0

Divide both sides of the equation by 4.

a^{2}d^{2} + b^{2}c^{2} - 2abcd = 0

(ad – bc)^{2} = 0

or ad = bc

or \\frac{a}{b}) = \\frac{c}{d})