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

2 Answers
Nov 20, 2017

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

Nov 20, 2017

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