As I'm not even sure what type of problem this is, so I can't research it.


My Question:

Could I get an explanation (with example) of how to solve the above equation? Or direct me to a resource with the explanation (preferably Khan Academy)? I am not looking for the answer to this problem, just the way I would solve it.

What I've attempted:

  • I have tried making it a perfect square, but can't get past the odd numbers.
  • I have tried squaring everything, and using the quadratic formula, but it doesn't match the answer format.

Any help is greatly appreciated!

  • 1
    $\begingroup$ Substract 3 and square. And use the quadratic formula, one of the roots should work. $\endgroup$ – YoTengoUnLCD Feb 6 '16 at 22:56
  • $\begingroup$ You have directly a quadratic equation in $y=\sqrt x $ $\endgroup$ – Piquito Feb 6 '16 at 23:55

The equation is $$3+5\sqrt{x}=2x.$$ If we make a change of variable $t=\sqrt{x}$ then it becomes

$$3+5t=2t^2.$$ This is a quadratic equation that, probably, you know how to solve it. Once you have the solution, say $t_1,t_2$ you has $x_1=\sqrt{t_1}$ and $x_2=\sqrt{t_2}$ (if $t_i\ge 0$) as possible solutions of the original equation. Check if they are solutions and you are done.

So, for equations $ax+b\sqrt{x}+c=0$ the above change works. If you have other situations as $$\sqrt{x-3}+5=2x$$ isolate the root. That is, $$\sqrt{x-3}=2x-5$$ and square both sides of the equation. Solve it and check if the solutions solve the original equation. You need to check because when we square we have $(-2)^2=2^2$ but $-2\ne 2.$

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  • $\begingroup$ Darn you beat me to it! $\endgroup$ – Piotr Benedysiuk Feb 6 '16 at 23:01
  • $\begingroup$ Thank you! I couldn't have asked for a better explanation! $\endgroup$ – z7r1k3 Feb 6 '16 at 23:24
  • $\begingroup$ @Spencer4134 You're welcome. $\endgroup$ – mfl Feb 6 '16 at 23:26

$$3+5\sqrt x=2x\implies 5\sqrt x=2x-3\implies25x=4x^2-12x+9\implies$$


But since we squared we could have gotten some false solution (in fact we did). Check both suposed solutions in the original equation.

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  • $\begingroup$ +1 for recognising the false solution. $\endgroup$ – Mr Pie Jun 7 at 8:25

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