Help with pre-calc problem?

Question

OK, this is pissin me off...I've been at it for over 30 min and can't find anything in the book about it. I remember learning how to do this at some long lost point, but that was years ago. Can you tell me why I do what this does?

sqrt=square root
sq=squared

(4+sqrt(x-5))sq

Now I know when you take the square of a sqrt you get whatever was under the square root, but in this case you dont. I get something along the lines of 16 + 8(x-5) +xsqrt(x-5)

What the hell? I still have the square root? I'm so lost...and where did the 8 come in? ughh

In the answer there is no "x" in front of the value of xsqrt(x-5)...sorry for the confusion... still dont get it though

Answer 1

Treat this as any other quadratic such as (5+x)^2=25+10x+x^2
(4+sqrt(x-5))^2=16+8sqrt(x-5)+x-5
The 8 comes from the square double square rule that is used in all quadratics period. (n+y)^2=n^2+2ny+y^2

maussy is wrong because he/she forgot to add the 2ny to the answer and this is an expression therefore there is no exact value for x.

Because you have no equation you can't solve for x so all you can do is simplify and the answer is 11+x+8sqrt(x-5)

Answer 2

I try (4+sqr(x-5) )^2 = 16 +8sqr(x-5) +x-5

11 +x = -8sqr(x-5) square the two sides

121+22x+x^2 = 64 (x-5) = -320+64x

441-42x +x^2 =0 =(x-21)^2

so x=21

verify