Question

√3x+24=x+2

Answer 1

Square both sides
\[(\sqrt{3x+24})^{2} = (x+2)^{2}\]
\[3x+24 = x^{2}+4x+4\]

Answer 2

\[\large \sqrt{3x+24}=x+2\]



\[\large 3x+24=(x+2)^2\]



\[\large 3x+24=x^2+4x+4\]



\[\large 0=x^2+4x+4-3x-24\]



\[\large 0=x^2+x-20\]



\[\large (x+5)(x-4)=0\]



\[\large x+5=0 \ \ \textrm{or} \ \ x-4=0\]



\[\large x=-5 \ \ \textrm{or} \ \ x=4\]



If you plug in each possible soluton back into the original equation, you'll find that only x = 4 works.


So the only solution is x = 4.