Question

If the roots of the equation (a² + b²)x² − 2 (ac + bd)x + (c² + d²) = 0 are equal, prove that `a/b=c/d`.

Answer 1

The given quadric equation is (a² + b²)x² − 2 (ac + bd)x + (c² + d²) = 0, and roots are real

Then prove that `a/b=c/d`.

Here,

a = (a² + b²), b = -2 (ac + bd) and c = (c² + d²)

As we know that D = b² - 4ac

Putting the value of a = (a² + b²), b = -2 (ac + bd) and c = (c² + d²)

D = b² - 4ac

= {-2(ac + bd)}² - 4 x (a² + b²) x (c² + d²)

= 4(a²c² + 2abcd + b² + d²) - 4(a²c² + a²d² + b²c² + b²d²)

= 4a²c² + 8abcd + 4b²d² - 4a²c² - 4a²d² - 4b²c² - 4b²d²

= -4a²d² - 4b²c² + 8abcd

= -4(a²d² + b²c² - 2abcd)

The given equation will have real roots, if D = 0

-4(a²d² + b²c² - 2abcd) = 0

a²d² + b²c² - 2abcd = 0

(ad)² + (bc)² - 2(ad)(bc) = 0

(ad - bc)² = 0

Square root both sides we get,

ad - bc = 0

ad = bc

`a/b=c/d`

Hence `a/b=c/d`