Question

X^2+10x-7 / x^2+10x+25=0 how do i solve for x?

Answer 1

Factorize both the numerator and denominator then equate to zero ...you'll get X

Answer 2

You just have to solve the next equation:
x²+10x-7=0
Because, if the numerator is equal to "0", then the next expression
(x²+10x-7) /( x²+10x+25)=0

Therefore:
x²+10x-7=0
x=[-10⁺₋√(100+28)]/2=(-10⁺₋√128)/2=(-10⁺₋8√2)/2=-5⁺₋4√2
We have two possible solutions:

x₁=-5-4√2
x₂=-5+4√2

Now, we have to check these possible solutions.
Remember the denominator cannot be equal to "0".

if x=-5-4√2

[(-5-4√2)²+10(-5-4√2)-7] / [(-5-4√2)²+10(-5-4√2)+25]=0/32=0

Therefore: x=-5-4√2 is a solution.

if x=-5+4√2

[(-5+4√2)²+10(-5+4√2)-7] / [(-5+4√2)²+10(-5+4√2)+25]=0/32=0

Therefore: x=-5+4√√2 is a solution.

Answer: we have two solutions for x:
x₁=-5-4√2
x₂=-5+4√2