Question

sqrt 2x-1 + sqrt x-4 = 4

Answer 1

\[\sqrt{2x-1}+\sqrt{x-4}=4\]

Answer 2

We want to use the difference of two squares: (a+b)(a-b)=a^2-b^2. Multiply both sides by \[(\sqrt{2x-1}-\sqrt{x-4})\]And we get:
\[(\sqrt{2x-1}+\sqrt{x-4})(\sqrt{2x-1}-\sqrt{x-4})=4(\sqrt{2x-1}-\sqrt{x-4})\]
\[(2x-1)-(x-4)=4\sqrt{2x-1}-4\sqrt{x-4} \rightarrow x+3=4\sqrt{2x-1}-4\sqrt{x-4}\]
Now we have two equations which we can work with:\[\sqrt{2x-1}+\sqrt{x-4}=4\]\[4\sqrt{2x-1}-4\sqrt{x-4}=x+3\]
Add four times the first equation to the second equation and we get:\[8
\sqrt{2x-1}=x+19\]\[64(2x-1)=x^2+38x+361\]\[x^2-90x+425=0\]
This gives x=5, x=85.