help pleaseeee?

Question

1) solve: sq rt. of (x+6) = x-6
2) 1/(x^2-6x+5) = (1/(x+5)) + (8/(x^2-25))

Answer 1

Hi,
You did an excellent job of grouping your expressions. Thanks!
1) solve: sq rt. of (x+6) = x-6
√(x+6) = x-6
x+6 = (x-6)²….(Square both sides.)
x+6 = x²-12x +36
x²-13x +30 = 0
(x-3)(x-10) = 0
x-3=0
x=3
x-10=0
x=10
If you try both of these solutions in the original equation, you’ll see that only x=10 works. That’s because 3 was generated as an extraneous solution when we squared the terms.

2) 1/(x^2-6x+5) = (1/(x+5)) + (8/(x^2-25))
(1/((x-5(x-1)) =((1/(x+5)) + 8/((x-5)(x+5)
Multiply by the LCD of (x-5)(x-1)(x+5) to get the following:
X+5 =(x-5)(x-1) + 8(x-1)
X+5 =x²-6x+5 +8x-8
X²-x-8 = 0
If you plug that into the quadratic formula, you’ll get this:
X=.5+-.5√(33)
Since the quadratic formula is only arithmetic, I’ll leave that to you.

FE