Question

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

Answer 1

SOLUTION :  

Given : (x - a)/(x - b) + ( x - b)/( x - a) = a/b + b/a  

[(x - a)² + (x -b)² ] / [(x - a)(x - b)] = [(a² + b²)/ab]

[ By taking LCM]

(x² - 2a + a² + x² - 2b + b²) /(x² - ax - bx + ab) = [(a² + b²)/ab]

ab(x² - 2a + a² + x² - 2b + b²) = (a² + b²)(x² - ax - bx + ab)

ab(x² - 2a + a² + x² - 2b + b²) = (a² + b²)(x² - ax - bx + ab)

ab(x² + x² - 2a - 2b + a² + b²) = (a² + b²)(x² - ax - bx + ab)

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

[ 2abx² -2abx(a + b) + ab(a² + b²)] = (a² +b²)x² -(a² +b²)(a +b)x +(a² + b²)ab  

[2abx² - 2abx(a + b) + ab(a² + b²)] = (a² +b²)x² - (a² +b²)(a +b)x + (a² + b²)ab  

[2abx² - (a² +b²)x²  - 2abx(a + b) + (a² +b²)(a +b)x] = (a² + b²)ab - (a² + b²)ab  

[(2ab - a² - b²)x² - (a +b)(2ab - a² - b²)x = 0

(a² + b² -2ab)x² - (a + b)(a² + b² - 2ab)x = 0

(a - b)²x² - (a + b)(a - b)² x =0

[We know that , (a - b)² = a² + b² - 2ab]

(a - b)² [x² - (a + b) x ] = 0

x² - (a + b)x = 0

x[ x - (a + b)] =0

x = 0  or x - (a + b) = 0

x = 0  or  x = a + b

Hence, the roots of the quadratic equation (x - a)/(x - b) + ( x - b)/( x - a) = a/b + b/a  are 0 & (a + b) .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution :

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

=> (x-a)/(x-b) - a/b = b/a - (x-b)/(x-a)

=>[b(x-a)-a(x-b)]/[(x-b)b]=[b(x-a)-a(x-b)]/[a(x-a)]

=>[bx-ab-ax+ab]/[(x-b)b]=[bx-ab-ax+ab]/[a(x-a)]

=> ( bx-ax )/[(x-b)b] = ( bx-ax )/[a(x-a)]

=> [x(b-a)]/[(x-b)b]=[x(b-a)]/[a(x-a)]

=> [x(b-a)]/[(x-b)b]-[x(b-a)]/[a(x-a)]=0

=> x(b-a){ 1/[(x-b)b] - 1/[a(x-a)] = 0

=> x( a - b ) = 0 or {1/[(x-b)b]-1/[a(x-a)]=0

=> x = 0 or 1/[(x-b)b] = 1/[a(x-a)]

=> (x-b)b = a(x-a)

=> bx - b² = ax - a²

=> bx - ax = b² - a²

=> ( b - a )x = ( b + a )( b - a )

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

=> x = b + a

Therefore ,

x = 0 or x = a + b

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