Question

Solve the following quadratic equation by factorization: $\frac{a}{x-b}+\frac{b}{x-a}=2$

Answer 1

SOLUTION :  

Given : a/(x - b) + b/(x - a) = 2  

[a(x - a) + b(x - b)]/ (x - b) (x - a) = 2

[ By taking LCM]

[ax - a² + bx - b²]  / [x² - ax -  bx + ab] = 2

[ax - a² + bx - b²]  = 2  [x² - ax -  bx + ab]  

[ax - a² + bx - b²]  = 2 x² - 2ax -  2bx + 2ab

2x² - 2ax - 2bx + 2ab - ax + a² - bx + b² = 0

2x² - 2ax - ax - 2bx - bx + a² + b² + 2ab =0

2x² + x [ - 2a - a - 2b - b ] + a² + b² + 2ab =  

2x² + x [ - 3a  - 3b ] + a² + b² + 2ab = 0

2x² - 3x [ a  + b ] + ( a + b)² = 0

2x² - 2(a + b) x - (a + b) x +( a + b)² = 0

2x(x -(a + b) - (a + b) (x - (a + b) = 0

(2x - (a + b)) (x - (a + b) = 0

(2x - (a + b))  = 0  or  (x - (a + b) = 0

2x = (a + b)  or  x = (a + b)  

x = (a + b)/2  or x = (a + b)

Hence, the roots of the quadratic equation a/(x - b) + b/(x - a) = 2   are  (a + b)/2  & (a + b) .

★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION METHOD :  

We first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution :

Factorization method :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\:>2x = (a + b)\:or\: x = (a + b)$

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

Quadratic method ( not in question ) :

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

$a=-2$  ======= > ( 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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