Can't seem to solve a radical equation? Question is : $\sqrt{x+19} + \sqrt{x-2} = 7$ So there is this equation that I've been trying to solve but keep having trouble with.   
The unit is about solving Radical equations and the question says
Solve:
$$\sqrt{x+19} + \sqrt{x-2} = 7$$
I don't want the answer blurted, I want to know how it's done, including steps please.
Thank you!
 A: To solve a simple sum of radical equations, you can use with fun and profit the radical equation calculator. It gives you some steps, as describe below:

In your case, a sum of two radicals equating a constant, the "isolate one root and square" method can help:
$$\sqrt{x+19}+\sqrt{x−2}=7\,.$$
So:  isolate, as much as you can, one radical on one side
$$\sqrt{x+19}=7 - \sqrt{x−2}$$
square:
$$x+19 = 49 + x-2-14 \sqrt{x−2}$$
simplify and isolate one radical again
$$28 =14 \sqrt{x−2}$$
$$2=\sqrt{x−2}$$
then square again, and you get $x=6$. Of course, you need to verify that the solution is valid, for instance that it does not involve radicals of negative numbers.
This way, you minimize products of radicals (like $\sqrt{x+19}\sqrt{x−2}$) that are easy traps. This method would work more generally, provided there are solutions, for the generic equation:
$$\sqrt{a_1 x+a_2}+\sqrt{a_3 x+a_4}=a_5$$
which you can try to exercise your style. More references:


*

*Steps required for Solving Problems Containing One Square Root

*Steps required for Solving Problems Containing Two Square Roots

*How to Solve a Square Root Equation
A: $\sqrt{x+19} + \sqrt{x-2} = 7$
Squaring both sides, we have 
$x+19+2\sqrt{x+19}\sqrt{x-2}+x-2=49$
Collecting terms, we have 
$2x+17+2\sqrt{x^2+17x-38}=49$
$\sqrt{x^2+17x-38}=\dfrac{32-2x}{2}$
Squaring again
$x^2+17x-38=\dfrac{1024-128+4x^2}{4}$
$x^2+17x-38=256-32x+x^2$
$49x=294$
$\therefore x=\dfrac{294}{49}=6$ 
We can easily verify that this is a correct solution.
A: Multiply both sides by $\sqrt{x+19} -\sqrt{x-2} $ to get
$$x+19 -(x-2) = 7 (\sqrt{x+19} - \sqrt{x-2}) \\
21 = 7 (\sqrt{x+19} - \sqrt{x-2}) \\
3 =\sqrt{x+19} - \sqrt{x-2}$$
Adding this to the original equation you get
$$2\sqrt{x+19}=10 \Rightarrow x+19=25 \Rightarrow x=6$$
P.S. You can find the same method employed in my answer here:
to this similar question
A: Hope this acceptable...
$\sqrt{{x}+\mathrm{19}}+\sqrt{{x}−\mathrm{2}}=\mathrm{7} \\ $
$\sqrt{{x}+\mathrm{19}}=\mathrm{7}−\sqrt{{x}−\mathrm{2}} \\ $
${x}+\mathrm{19}=\mathrm{49}−\mathrm{14}\sqrt{{x}−\mathrm{2}}+\left({x}−\mathrm{2}\right) \\ $
$\mathrm{0}=\mathrm{28}−\mathrm{14}\sqrt{{x}−\mathrm{2}} \\ $
$\mathrm{2}=\sqrt{{x}−\mathrm{2}}\Rightarrow\mathrm{4}={x}−\mathrm{2}\Rightarrow{x}=\mathrm{6} \\ $
