Question

Find the value of x , (x-a/x-b)+(x-b/x-a) = (a^2+b^2/ab)

Answer 1

Hey i have used shortcut for this not sure about method

Answer 2

Given Equation is$\frac{x-a}{x-b} + \frac{x-b}{x-a} = \frac{ a^{2} + b^{2} }{ab}$ 

On cross multiplication we get

$\frac{x-a}{x-b}*ab(x-b)(x-a)+\frac{x-b}{x-a}*ab(x-b)(x-a)= \frac{a^{2}+ b^{2}}{ab}ab(x-b)(x-a)$


ab(x-a)^2 + ab(x-b)^2 = (a^2+b^2)(x-b)(x-a)


We know that (a-b)^2 = a^2+b^2-2ab.


ab(x^2+a^2-2ax) + ab(x^2+b^2-2bx) = (a^2+b^2)(x-b)(x-a)


abx^2 +a^3b - 2a^2bx + abx^2 + ab^3 - 2ab^2x = a^2x^2-a^3x-a^2bx+a^3b+b^2x^2-ab^2x-b^3x+ab^3


a^3b - 2a^2bx + 2abx^2 - 2ab^2x +b^3x - b^2x^2  = a^2x^2 - a^3x - a^2bx + a^3b


b^3x - ab^2x -b^2x^2 - 2a^2bx + 2abx^2 = a^2x^2 - a^3x - a^2bx 


-b^2x^2 + 2abx^2 + b^3x - ab^2x - a^2bx  + a^3x = a^2x^2


(-b^2+2ab-a^2)x^2 + (b^3 -ab^2 -a^2b+a^3)x = 0

We know that Quadratic Equation formula is 

$\frac{-b+ \sqrt{b^2 - 4ac} }{2a}$

= $\frac{-(b^3-ab^2-a^2b+a^3)+ \sqrt{(b^3-ab^2-a^2b+a^2)-4(-b^2+ 2ab-a^2)*0} }{2(-b^2+2ab-a^2)}$


= $\frac{-(b^3-ab^2-a^2b+a^3) + a^3-a^2b-ab^2+b^3}{2(-a^2+2ab-b^2)}$

= $\frac{-b^3+ab^2+a^2b-a^3 + a^3-a^2b-ab^2+b^3}{2(-a^2+2ab-b^2)}$

= $\frac{0}{2(-a^2+2ab-b^2)}$

= 0.


If u solve this equation using  $\frac{-b - \sqrt{b^2-4ac}}{2a}$, U will get.

$- \frac{b^3-ab^2-a^2b+a^3}{-b^2+2ab-a^2}$


Therefore the value of x = 0 and - $\frac{b^3-ab^2-a^2b+a^3}{-b^2+2ab-a^2}$


Hope this helps!