Question

If the roots of the Quadratic equation (a²+b²)x²-2 (ac+bd)x+c²+d²=0 are
equal Prove that a/b =c/d​

Answer 1

Answer:

a/b = c/d

Step-by-step explanation:

Roots of (a²+b²)x²-2(ac+bd)x+c²+d²=0 are equal, which means discriminant is 0.

= > discriminant = 0

= > [-2( ac + bd )]² - 4[(a² + b²)(c² + d²) = 0

= > [4(a²c² + b²d² + 2abcd)] - 4[ a²c² + a²d² + b²c² + b²d²] = 0

= > 4[ a²c² + b²d² + 2abcd - a²c² - a²d² - b²c² - b²d²] = 0

= > 4[ + 2abcd - a²d² - b²c²] = 0

= > 2abcd - a²d² - b²c² = 0

= > a²d² + b²c² - 2abcd = 0

= > ( ad - bc )² = 0

= > ad - bc = 0

= > ad = bc

= > a/b = c/d

As desired

Answer 2

Given :

The quadratic equation is :

•(a² + b²)x² -2(ac+bd)x + c²+d²=0

•The roots of this quadratic equation are equal.

Solution :

We know that , if the roots of a quadratic equation are equal then the discriminant , b² - 4ac=0 .

Therefore we have ,

$\sf \implies \{ -2(ac+bd) \}^{2} -4(a^{2} + b^{2} )(c^{2} + d^{2} ) = 0 \\\\ \sf \implies 4(ac+bd)^{2} - 4(a^{2} + b^{2})(c^{2} + d^{2} ) = 0 \\\\ \sf \implies 4(a^{2}c^{2} + b^{2}d^{2} + 2abcd) = 4(a^{2}c^{2} + a^{2}d^{2} +b^{2}c^{2}+ b^{2}d^{2} ) \\\\ \sf \implies a^{2}c^{2} + b^{2}d^{2} + 2abcd = a^{2}c^{2} + b^{2}d^{2} + a^{2}d^{2}+b^{2}c^{2} \\\\ \sf \implies 2abcd = a^{2}d^{2} + b^{2}c^{2} \\\\ \sf \implies a^{2}d^{2} + b^{2}c^{2} - 2abcd = 0 \\\\ \sf \implies (ad - bc)^{2} = 0 \\\\ \sf \implies ad -bc = 0 \\\\ \sf \implies ad = bc \\\\ \sf \implies \dfrac{a}{b} = \dfrac{b}{c}$

Hence proved