Question

How to solve using quadratic formula: `(b/(x-a))+(a/(x-b)=2` ?

Answer 1

Multiply by x-a and x-b

Explanation:

Take note of the following before even using the quadratic formula:
x=a+b is one answer, as each fraction becomes b/b+a/a, which is two.

Step by step:

`(b/(x-a))+(a/(x-b))=2`
Multiply by x-a
`b+((a(x-a))/(x-b))=2(x-a)`
Multiply by x-b
`b(x-b)+a(x-a)=2(x-a)(x-b)`
Expand
`bx-b^2+ax-a^2=2x^2-2ax-2bx+2ab`
Collect like terms
`0=2x^2-3(a+b)x+(2ab+a^2+b^2)`
`0=2x^2-3(a+b)x+(a+b)^2`
There's a regular quadratic for you to solve using the quadratic formula.
for `dx^2+ex+f`, the solutions are `x=(-e+-sqrt(e^2-4df))/(2d)`
The solutions gotten:
`x=(3(a+b)+-sqrt(3^2(a+b)^2-4*2*(a+b)^2))/(2*2)`
`x=(3(a+b)+-sqrt(9(a+b)^2-8(a+b)^2))/(4)`
`x=(3(a+b)+-(a+b))/(4)`
The two solutions are therefore:
`x=(a+b)/2,a+b`

Answer 2

`x_1=(a+b)/2` and `x_2=a+b`

Explanation:

`b/(x-a)+a/(x-b)=2`

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

`(xb-b^2+ax-a^2)/[x^2-(a+b)*x+ab]=2`

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

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

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

`[x-(a+b)]*[2x-(a+b)]=0`

Hence `x_1=(a+b)/2` and `x_2=a+b`