Can someone help me solve these equatiions

[SQRT(4 - x)] - [SQRT(x + 6)] = 2

4 answers

[SQRT(4 - x)] - [SQRT(x + 6)] = 2
√(4-x) = 2 + √(x+6)
square both sides
4-x = 4 + 4√(x+6) + x+6
-2x - 6 = 4√(x+6)
-x - 3 = 2√(x+6)
square again

x^2 + 6x + 9 = 4(x+6)
x^2 + 2x - 15 = 0
(x+5)(x-3) = 0

x = -5 or x = 3

since we squared both answers must be verifies
if x=-5
LS = √9 - √1
= 3 - 1
which is not the RS, so x=-5 does NOT work

if x= 3
LS = √1 - √9
= 1-3
= -2, which is not the RS

So, there is NO solution

if x=
I don't see what -5 does not work. The right side is 2

If you admit negative square roots, +3 works also
"what" should be "why" in my previous answer. I often can't get either the numbers or the words right.
of course you are right, x = -5 works.

don't know what I was thinking!
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