# How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, 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 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.

• Square both sides, then isolate the square root and square again. Jul 18, 2011 at 4:22
• ....and check for extraneous roots. There's always a danger of those when you square both sides. Jul 18, 2011 at 4:24
• You guys are too fast! :) Jul 18, 2011 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, 2011 at 4:28
• @barf, why does $x$ need to be an integer? Jul 18, 2011 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.

• Perhaps slightly easier than that is to make the substitution $x = t+4$ before moving over the $\sqrt{6x-17}$ term over. Jul 18, 2011 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 $x=t+3$...) Jul 18, 2011 at 5:16
• @Aryabhata: Ahh, okay, I can see the motivation in the $5x-20$, then. To me, that (or actually any substitution) seems like a tradeoff of risk of error in additional algebraic steps (from the substitution) against risk of error in computation with messy numbers (that resulted from not substituting). Jul 18, 2011 at 5:20
• 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. Jul 18, 2011 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, 2011 at 12:23

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. $x=79/25$).

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...

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.

• -17 not -7 in the 2nd line Jul 18, 2011 at 4:32
• @Zarrax: Thanks. Fixed, and in the third. Jul 18, 2011 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.

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

Let's begin by adding $\sqrt{6x-17}$ to both sides of the equation:

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

Then, we can square both sides:

$\big(4\sqrt{x-3}\big)^{2}=\big(3+\sqrt{6x-17}\big)^{2}$

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

$\Rightarrow 16x-48=9+6\sqrt{6x-17}+6x-17$

$\Rightarrow 10x-40=6\sqrt{6x-17}$

The equation can be simplified by dividing both sides by $2$:

$\Rightarrow 5x-20=3\sqrt{6x-17}$

Now, we can square both sides of the equation once again to remove the square root sign:

$\Rightarrow (5x-20)^{2}=\big(3\sqrt{6x-17}\big)^{2}$

$\Rightarrow 25x^{2}-200x+400=9(6x-17)$

$\Rightarrow 25x^{2}-200x+400=54x-153$

$\Rightarrow 25x^{2}-254x+553=0$

Now, we solve for $x$ using the quadratic formula:

$\Rightarrow x=\dfrac{-(-254)\pm\sqrt{(-254)^{2}-4(25)(553)}}{2(25)}$

$\hspace{9 mm}=\dfrac{254\pm96}{50}$

$\hspace{9 mm}=\dfrac{127\pm48}{25}$

$\hspace{9 mm}=\dfrac{79}{25}$,$\hspace{1 mm}7$

However, $x=\dfrac{79}{25}$ does not work in the original equation.

Therefore, the solution to the equation is $x=7$.

• $x=\frac{79}{25}$ doesn't work in the original equation. Jun 14, 2016 at 21:11
• Thanks for pointing that out Jun 14, 2016 at 21:21

There is a trick for problems like this that I really love to do.

\begin{align} 4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(multiply)} \\ \text{Let } 4\sqrt{x-3} + \sqrt{6x-17} &= y \\ \hline 16(x-3)-(6x-17) &= 3y \\ 16x-48 - 6x + 17 &= 3y \\ y &= \dfrac{10x -31}{3} \\ \hline 4\sqrt{x-3} + \sqrt{6x-17} &= \dfrac{10x -31}{3} &\text{(so now we have this)} \\ 4\sqrt{x-3} - \sqrt{6x-17} &= 3 &\text{(subtract them)}\\ \hline 2\sqrt{6x-17} &= \dfrac{10x -40}{3} \\ \sqrt{6x-17} &= \dfrac{5x - 20}{3} \\ 3\sqrt{6x-17} &= 5x - 20 \\ 54x - 153 &= 25x^2 - 200x + 400 \\ 25x^2 - 254x + 553 &= 0 \\ (x-7)(25x - 79) &= 0 \\ x &\in \left\{ 7, \dfrac{79}{25}\right\} \end{align}

Check for extraneous roots.

When $$x = 7$$ $$(4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot 2 - 5 = 3$$

When $$x = \dfrac{79}{25}$$ $$(4\sqrt{x-3} - \sqrt{6x-17})= 4\cdot \dfrac 25 - \dfrac 75 = \color{red}{\frac 15}$$

So we reject $$x = \dfrac{79}{25}$$ as an extraneous root and accept $$x=7$$,

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.

• that doesn’t actually help you find the point though. You still have to guess or to solve it. More useful if guessing because you can know you only need one answer, but not really in the spirit of OP asking how to solve it Jun 14, 2016 at 21:32
• @BrevanEllefsen You are right. This is not a generalized method. It just allows to save a lot of time in case you can guess roots from a graph. It does solve this particular equation. This is a good approach in exam situation for example. Jun 15, 2016 at 9:54

There are more creative subsitutions. For the square roots to be defined, $$x ≥ 3$$ and $$x ≥ \frac{17}{6}$$, so the substitution $$x = 3 \sec^2 u$$ is valid as its range is $$[3, \infty)$$, so for every value of $$x$$, there is a corresponding value of $$u$$. Doing this gives:

$$4 \sqrt{3(\sec^2 u - 1)} - \sqrt{18 \sec^2 u - 17} = 3$$ $$\Rightarrow - \sqrt{18 \sec^2 u - 17} = 3- 4 \sqrt{3} \tan u$$ $$\Rightarrow 18 \sec^2 u - 17 = 9- 24 \sqrt{3} \tan u+48\tan^2 u$$

$$\Rightarrow 18(\tan^2 u+1) - 17 = 9 - 24 \sqrt{3} \tan u+ 48 \tan^2 u$$ $$\Rightarrow 30 \tan^2 u- 24 \sqrt{3} \tan u+8=0$$ $$\Rightarrow 15 \tan^2 u - 12 \sqrt{3} \tan u + 4 = 0$$

and now using the quadratic formula, $$\tan u = \frac{2}{5 \sqrt3}, \frac{2}{\sqrt3}$$, so $$x = 3 \sec^2 u = 3(\tan^2u+1) = \frac{79}{25}, 7$$. However, substituting back into the original equation, $$x = \frac{79}{25}$$ does not work but $$x = 7$$ does. Hence $$x = 7$$ only.