How could I find the value of $x$ without squaring both sides given the equation $\sqrt{8} + \sqrt{18} = \sqrt{x}$?

Question

Our class got a challenge question on a recent test, that no one has been able to figure out. We know $x = 50$, but everyone seems to get stuck when they simplify it to $5\sqrt{2} = \sqrt{x}$. Any help? Is this even possible?

Answer 1

Let $x = 2y$.

Then $\sqrt{2*4} + \sqrt{2*9} = \sqrt{2y}$

So $\sqrt 2(\sqrt 4 + \sqrt 9) = \sqrt 2\sqrt y$

$2+3 =\sqrt y$

$5 = \sqrt y$

$y = 25$

$x = 50$.

Answer 2

You can just square both sides immediately, or you can simplify before squaring:

$$ \sqrt{8} + \sqrt{18} = \sqrt{x} \\ \sqrt{2\cdot 4} + \sqrt{2\cdot 9} = \sqrt{x} \\ 2\sqrt{2} + 3\sqrt{2} = \sqrt{x} \\ 5\sqrt{2} = \sqrt{x} \\ (5\sqrt{2})^2 = (\sqrt{x})^2 \\ 5^2\cdot \sqrt{2}^2 = x \\ 25\cdot 2 = x \\ 50 = x $$

Answer 3

Hint

$$5\sqrt{2}=\sqrt{x}\to \sqrt{50}=\sqrt{x}$$

the function $f(x)=\sqrt{x}$ is injective for $x\ge0$.

Can you finish?