Question

solve this question
use quadriatic formula for finding the zeroes

Answer 1

Solution :          

$\frac{a}{x-b}+\frac{b}{x-a}=2$

$=\:>\frac{a(x-a)+b(x-b)}{(x-a)(x-b)}=2$

$=\:> a(x-a)+b(x-b)=2(x-a)(x-b)$

$=\:>ax-a^2+bx-b^2=2(x^2-ax-bx+ab)$

$=\:>ax-a^2+bx-b^2=2x^2-2ax-2bx+2ab$

$=\:> 3ax+3bx-2x^2-2ab-a^2-b^2=0$

$=\:>x(3a+3b)-2x^2-(a^2+b^2+2ab)=0$

$=\:>-2x^2+x(3a+3b)-(a^2+ab+ab+b^2=0$

$=\:> -2x^2+x(3a+3b)-(a(a+b)+b(a+b))=0$

$=\:> -2x^2+x(3a+3b)-(a+b)(a+b)=0$

$\textsf{Comparing with }\mathsf{ax^2+bx+c=0\:we\:get:}$

$a=-2$  ======= > ( 1 )

$b=(3a+3b)$  ======= > ( 2 )

$c=-(a+b)(a+b)$  ======= > ( 3 )

$\triangle=b^2-4ac$

$=\:> (3a+3b)^2-4\times-(a+b)(a+b)\times(-2)$

$=\:> 9a^2+9b^2+18ab-8(a+b)^2$

$=\:> 9a^2+9b^2+18ab-8(a^2+b^2+2ab)$

$=\:>9a^2+9b^2+18ab-(8a^2+8b^2+16ab)$

$=\:> 9a^2+9b^2+18ab-8a^2-8b^2-16ab$

$=\:> a^2+b^2+2ab$

$=\:> a^2+ab+ab+b^2$

$=\:> a(a+b)+b(a+b)$

$=\:>(a+b)(a+b)$

$=\:> (a+b)^2$ ======= > ( 4 )

Quadratic Formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-b\pm\sqrt{\triangle}}{2a}$

From ( 4 ) :

$=\:> \frac{-b\pm\sqrt{(a+b)^2}}{2a}$

$=\:> \frac{-b\pm(a+b)}{2a}$

From ( 1 ) and ( 2 ) :

$=\:> \frac{-(3a+3b)\pm(a+b)}{2\times(-2)}$

$=\:> \frac{-3a-3b\pm(a+b)}{-4}$

Either :

$=\:> \frac{-3a-3b+a+b}{-4}$

$=\:> \frac{-2a-2b}{-4}$

$=\:> \frac{(-2)(a+b)}{-4}$

$=\:> \frac{a+b}{2}$

Or:

$=\:> \frac{-3a-3b-(a+b)}{-4}$

$=\:> \frac{-4a-4b}{-4}$

$=\:> \frac{-4(a+b)}{-4}$

$=\:> a+b$

ANSWER :

The zeroes of the quadratic equation are :

$\mathsf{Either\:\:\boxed{\frac{a+b}{2}}}\\\\\\\mathsf{Or\:\:\boxed{a+b}}$

Hope its helpful !

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