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Again a root problem.. $\sqrt{2x+5}+\sqrt{5x+6}=\sqrt{12x+25}$

Isn't there any standardized way to solve root problems..Can u plz help by giving some tips and stategies for root problems??

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Square and keep the radicals at one side, then square to find the roots, check for the extraneous roots possibly introduced by squaring – lab bhattacharjee Jul 22 '13 at 17:40
I don't see how this is "quadratics"? – user67258 Jul 22 '13 at 17:42
square both sides – eccstartup Jul 22 '13 at 17:42
@DannyCheuk You get a quadratic equation at the end. – Mlazhinka Shung Gronzalez LeWy Jul 22 '13 at 17:42
If I saw this problem, I would just test a few small integer values... $x=2$ works. (Obviously this isn't a rigorous method.) – The Chaz 2.0 Jul 22 '13 at 18:09

Square to get $$2x+5+2\sqrt{(2x+5)(5x+6)}+5x+6=12x+25$$

This reduces to $$2\sqrt{(2x+5)(5x+6)}=5x+14$$

Now square again, solve the quadratic, and check the solutions in the original equation. It doesn't get that unwieldy, and there is a solution hidden quite close to the surface - I found the formulation of the problem suggestive.

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i have been able to solve with this method ..but i am asking and neter or better aproach.. – maths lover Jul 22 '13 at 17:55
Well, each side of the equation is increasing with $x$ - one side increasing faster than the other. So there is at most one solution. When $x=0$ the left-hand side is less than the right hand side. When $x$ is very large it is greater. So if you spot a solution, that has to be the one, or try the substitution $x=y+2$ and solve for $y$. But nailing all that requires either solving the equation to get the right trick, or doing more work than you need to. – Mark Bennet Jul 22 '13 at 18:04

There isn't much you can do which is applicable to every problem of the form $$\sqrt{ax + b} + \sqrt{cx + d} = \sqrt{ex + f} $$ But, if you are lucky enough to have an equation where there exists $p,q$ such that$(ax+b)*(cx+d) = (qx + p)^2$ then this problem reduces quite nicely. You'd find that $$ex + f = (a + c + 2q)x + (b + d + 2p)$$

Of course, extraneous solutions will need to be considered, but .. I think you get my point

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