Question

How do you solve `(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)` and check for extraneous solutions?

Answer 1

`x=(7+-sqrt1401)/4`

Explanation:

As `(x-4)(x-3)=x^2-7x+12` - note that as they are in denominator `x!=3` and `x!=4`

`(x-7)/(x-4)+(x+8)/(x-3)=(x+158)/(x^2-7x+12)`

`hArr((x-7)(x-3)+(x+8)(x-4))/(x^2-7x+12)=(x+158)/(x^2-7x+12)`

or `((x-7)(x-3)+(x+8)(x-4))=(x+158)`

or `x^2-10x+21+x^2+4x-32=x+158`

or `2x^2-6x-11-x-158=0`

or `2x^2-7x-169=0`

and using quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)` as olution for quadratic equation `ax^2+bx+c=0`

`x=(-(-7)+-sqrt((-7)^2-4*2^*(-169)))/4`

= `(7+-sqrt(49+1352))/4=(7+-sqrt1401)/4`