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.
 A: 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...
A: 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.
A: 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.
A: 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 $x=7$.
A: 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.
A: 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. $x=79/25$).
A: Do not square!


*

*Move one root over equality sign. 

*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.
A: 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.
A: 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 $x=7$,
