Question

I need help with solving radical equations.
sqrt(8x-23)=4-x

Answer 1

squaring on both sides x^-8x+16=8x-23
x^-8x+16-8x+23=0
x^-16x+39=0

Answer 2

any restrictions?

Answer 3

\[8x-23\geq0\quad\wedge \quad 4-x\geq0\]\[x\geq23/8\quad\wedge \quad x\leq4\]

Answer 4

I need to solve for x the answer is 3, but I don't know how they got it. :(

Answer 5

\[\sqrt{8x-23}=4-x\]\[(\sqrt{8x-23})^2=(4-x)^2\]\[8x-23=16-8x+x^2\]\[0=39-16x+x^2\]\[0=(13-x)(3-x)\quad\Rightarrow\quad x=3\quad,\quad x\neq 13\,(13\not\in[23/8,4])\]