$\sqrt{3x}+7=x+1$

When faced with a problem where you have to solve for “x” but aren’t able to get rid of the $\sqrt{3x}$ by squaring both sides right away, you first have to isolate both “x” variables on one side of the equation.

Now we can get rid of the square root by squaring both sides.

$(\sqrt{3x})^2=(x-6)^2$

$3x=x^2-12x+36$

On the right side of the equation, $x-6$ turns into $x^2-12x+36$ because we have to expand. You aren’t able to just do the math.

Now, since our equation is quadratic we can make our equation equal to zero, and factor it to find our roots.

$0=x^2-15x+36$

Lucky for us this equation factors nicely:

\$latex 0=(x-12)(x-3)

So, x=12 and/or x=3

To check we plug one of our newly found “x’s” into our equation. If one doesn’t work, there’s still a chance that the other one will. ALWAYS check both!

$\sqrt{3x}+7=x+1$

$\sqrt{3*3}+7=3+1$

$\sqrt{9}+7=4$

$3+7=4$

$10=4$

This one DOES NOT work!!! However, we’ll check x=12 to check if that one does work.

$\sqrt{3x}+7=x+1$

$\sqrt{3*12}+7=12+1$

$\sqrt{36}+7=13$

$6+7=13$

$13=13$

So x=12 is a root, but x=3 is not. Therefore, the equation only has one root.