Question

How do you solve `10/(x^2-25) - 1/10 = 1/(x - 5)`?

Answer 1

`x_1=5`
`x_2=-15`

Explanation:

`x^2-25=(x-5)*(x+5)` so we have

`10/((x+5)*(x-5))-1/10=1/(x-5)`

multiply both sides by `10(x+5)(x-5)`

`10(x+5)(x-5)10/[(x+5)(x-5)]-10(x+5)(x-5)1/10=10(x+5)(x-5)1/(x-5)`

`10color(red)cancel(x+5)color(red)cancel(x-5)10/[color(red)cancel(x+5)color(red)cancel(x-5)]-color(blue)cancel10(x+5)(x-5)1/color(blue)cancel10=10(x+5)color(green)cancel(x-5)1/color(green)cancel(x-5)`

`100-(x^2-25)=10x+50`

`100-x^2+25-10x-50=0`

`-x^2-10x+75=0`

`x^2+10x-75=0`

now use the formula `x_(1//2)=(-b+-sqrt(b^2-4ac))/(2a)`
`a=1`
`b=10`
`c=-75`

`x_(1//2)=(-10+-sqrt(100-4(-75)))/(2)`
`x_(1//2)=(-10+-sqrt(400))/(2)`
`x_(1//2)=(-10+-20)/2`
`x_1=(-10+20)/2=10/2=5`
`x_2=(-10-20)/2=-30/2=-15`