# How to solve this equation with two square root terms?

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 embarassed 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

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Square both sides, then isolate the square root and square again. –  Qiaochu Yuan Jul 18 '11 at 4:22
....and check for extraneous roots. There's always a danger of those when you square both sides. –  Michael Hardy Jul 18 '11 at 4:24
You guys are too fast! :) –  Bruno Joyal Jul 18 '11 at 4:24
I think this problem is actually a perfect candidate for guess-and-check. Without being quantitative, you need x to be reasonably small, because $$4\sqrt{x-3} > \sqrt6\sqrt{x-17/6},$$ so the expression on the left hand side is increasing as $x$ increases. You also need $x$ to be greater than or equal to 3, so you dont get an imaginary number, and of course $x$ needs to be an integer. From there you only need to guess a very few til you get $x = 7$ as desired. Of course everyone else is right, but my point is that guessing is quick and useful sometimes! (e.g., for finding antiderivatives...) –  barf Jul 18 '11 at 4:28
@barf, why does $x$ need to be an integer? –  Gerry Myerson Jul 18 '11 at 4:34

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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Perhaps slightly easier than that is to make the substitution $x = t+4$ before moving over the $\sqrt{6x-17}$ term over. –  Aryabhata Jul 18 '11 at 5:15
@Aryabhata: Maybe it's just that I'm up past my bedtime, but I'm not seeing it—why $x=t+4$? (I could see maybe $x=t+3$...) –  Isaac Jul 18 '11 at 5:16
Then you get $5t = 3 \sqrt{6t + 7}$ and if you divide by $t$, you get $5 = 3\sqrt{6/t + 7/t^2}$ and squaring gives a quadratic in $1/t$. –  Aryabhata Jul 18 '11 at 5:18
Yeah, the comment was in the same spirit as your answer and I suppose the idea can be generalized to equations of the form $A\sqrt{ax+b} + B\sqrt{cx + d} = C$, making me think it might be worth mentioning. –  Aryabhata Jul 18 '11 at 6:12
Another reason to favor moving-things-around to squaring-both-sides as an initial step: Consider the case of three (or more) square roots on one side of the equation. –  Blue Jul 18 '11 at 12:23

You can square the equation and you end up with $$16(x-3)+(6x-17)-9=8\sqrt{(x-3)(6x-17)}$$ i.e. $$22x-74=8\sqrt{(x-3)(6x-17)}$$ and square again to obtain $$484x^2-44\cdot 74 x+74^2=64\cdot(x-3)(6x-17)$$ which is a second degree equation in $x$ you know how to solve! This yields at most $2$ solutions to your problem, and you can check by hand if both solutions work or if one doesn't work for instance because the term inside the square root is negative...

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If you want less square roots, you may try replacing $x=t^2+3$ to simplify the equation to get $$4t-\sqrt{6t^2+1}=3$$then$$(4t-3)^2=6t^2+1$$and finally$$5t^2-12t+4=0$$which gives you $t=2$ (i.e. $x=7$) or $t=2/5$ (i.e. $x=79/25$).

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Squaring both sides is a good first step, as it reduces the number of square roots to one. Another square will fix that.

$4\sqrt{x-3}-\sqrt{6x-17}=3$

$16(x-3)+8\sqrt{(x-3)(6x-17)}+6x-7=9$

$8\sqrt{(x-3)(6x-17)}=64-22x$

and another square gets rid of the square root, leaving a quadratic. Make sure to check the roots back into the original equation, as squaring is not reversible.

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-17 not -7 in the 2nd line –  Zarrax Jul 18 '11 at 4:32
@Zarrax: Thanks. Fixed, and in the third. –  Ross Millikan Jul 18 '11 at 4:39

If you square each side, you end up with only one square root term in the resulting expression. Isolate it on one side and square again, you'll get a quadratic in $x$ which you can solve.

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Do not square!

1. Move one root over equality sign.
2. A strictly increasing and a strictly decreasing functions can cross at no more than 1 point. It is easy to prove: 2nd crossing point will violate condition of functions being strictly increasing / decreasing.

Here is my answer to same question.

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