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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.

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    Haha Bruno, those kind of questions get ripped off by the first guy who sees them.2011-07-18

7 Answers 7

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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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    @Blue Ugh, that sounds hoooorrribbblleeee! Squaring everything with $n$ square root results in a polynomial of degree $2^{n-1}$, something I don't want to look at.2016-06-14
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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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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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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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    @Zarrax: Thanks. Fixed, and in the third.2011-07-18
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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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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 $x=7$.

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    Thanks for pointing that out2016-06-14
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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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    @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.2016-06-15