#### 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 `a/b=c/d`.

#### Solution

The given quadric equation is (a^{2} + b^{2})x^{2} − 2 (ac + bd)x + (c^{2} + d^{2}) = 0, and roots are real

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

Here,

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

As we know that D = b^{2} - 4ac

Putting the value of a = (a^{2} + b^{2}), b = -2 (ac + bd) and c = (c^{2} + d^{2})

D = b^{2} - 4ac

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

= 4(a^{2}c^{2} + 2abcd + b^{2} + d^{2}) - 4(a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2})

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

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

= -4(a^{2}d^{2} + b^{2}c^{2} - 2abcd)

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

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

a^{2}d^{2} + b^{2}c^{2} - 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`