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√3x+24=x+2
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Square both sides \[(\sqrt{3x+24})^{2} = (x+2)^{2}\] \[3x+24 = x^{2}+4x+4\]
\[\large \sqrt{3x+24}=x+2\] \[\large 3x+24=(x+2)^2\] \[\large 3x+24=x^2+4x+4\] \[\large 0=x^2+4x+4-3x-24\] \[\large 0=x^2+x-20\] \[\large (x+5)(x-4)=0\] \[\large x+5=0 \ \ \textrm{or} \ \ x-4=0\] \[\large x=-5 \ \ \textrm{or} \ \ x=4\] 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.
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