I am 42 years young and trying to brush up my 25 year old shelved algebra knowledge.

I am stumped with this problem.
I have the answer, but I don't know how to get there exactly. I know I will have to square both sides and that it will wind up as a Quad equation, but I am sooo lost on how to get there.

$\displaystyle 4\sqrt{x-1}+1=x$

2. 4rt(x-1)=x-1

squaring gives

16(x-1)=(x-1)(x-1)

gives x^2 -18x +17

factorise gives x=17 x=1

hope this helps.

3. let $\displaystyle x\ge1,$ thus $\displaystyle 4\sqrt{x-1}=x-1,$ which is equal to $\displaystyle \sqrt{x-1}\left( 4-\sqrt{x-1} \right)=0$ and from here it's $\displaystyle \sqrt{x-1}=0\implies x=1$ or $\displaystyle \sqrt{x-1}=4\implies x=17.$

4. Thank you, thank you! I was missing the swing of the 16 and 16x!
I have finished the problem!

Thank you again, both of you!