Algebra Precalculus – How to Solve for x in Equation with Two Square Root Terms

algebra-precalculus

So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were doing problems, we came about one that I have no idea how to solve. It seems to be a straight forward "here's an equation, solve for $x$." Just one problem, I ended up not being able to solve for $x$, making me feel embarrassed since I was trying to help her and I am supposed to be the one who is "good at math." Anyway, it would be a grreat help if anyone can help me out on this. The equation is….

$$4\sqrt{x-3} – \sqrt{6x-17} = 3$$

The answer comes out to be $x = 7$ because I did it on my calculator. I first tried to just square each side to get rid of the square roots. But once I "FOILed" the left side, there were still square roots and things didn't look good for me. By the way, I am sorry I don't know how to format this equation correctly.

Best Answer

You and every other answer thus far have talked about starting by squaring both sides, which works, but there is a slightly easier path to take. Rewrite your equation so that there is only one square root on each side, for example: $$4\sqrt{x-3}=3+\sqrt{6x-17}$$ Now, when you square both sides, you'll still have another square root to deal with, but it's not the square root of a product, like the $\sqrt{(x-3)(6x-17)}$ you would have had: $$16(x-3)=9+6\sqrt{6x-17}+(6x-17)$$ Expand, collect like terms, and rearrange to get the square root by itself: $$5x-20=3\sqrt{6x-17}$$ Square both sides again: $$25x^2-200x+400=9(6x-17)$$ $$25x^2-254x+553=0$$ Solving this gives $$x=7\text{ or }x=\frac{79}{25}$$ but $x=\frac{79}{25}$ doesn't work in the original equation, so $x=7$ is the only solution to the original equation.

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