Question

If the root of the equation (a^+ b^)x^-2 (ac+ bd )x+ (c^+d^)=0 are equal, prove that ab= cd

Answer 1

Answer:

ad = bc

Step-by-step explanation:

The given equation is-

$(a^{2}+b^{2})x^{2} -2(ac+bd)x + (c^{2}+d^{2}) = 0$

Comparing with general equation-

$Ax^{2}+Bx + C = 0$

Here, $A = a^{2}+b^{2}$

$B = -2(ac+bd)$ and

$C = (c^{2}+d^{2})$

We know that-

A quadratic equation has equals root if and only if-

$B^{2} - 4AC = 0$

$B^{2} = 4AC$

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

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

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

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

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

We know that-

$x^{2} + y^{2} - 2xy = (x-y)^{2}$

$(2ad)^{2} + (2bc)^{2} - 2(2ad)(2bc) = 0$

So, $(2ad - 2bc)^{2} = 0$

$2ad - 2bc = 0$

2ad = 2bc

ad = bc

             Hence proved.