Solve the expression below WITHOUT squaring both sides.

Question

isaiah.feynman · Answer

$$5-\frac{ 2 }{ x } = \sqrt{5-\frac{ 2 }{ x }}$$

terenzreignz · Answer

There are quite a few numbers which are their own square root.

terenzreignz · Answer

Only two of them, unless I'm mistaken.

isaiah.feynman · Answer

What are those numbers?

terenzreignz · Answer

Well, you may not be allowed to square both sides in that equation, but take a look at this and see if it makes things clearer:
$$\Large y = \sqrt y$$
This isn't your equation, so feel free to square both sides and solve for the two possible values of y.

anonymous · Answer

I got that there's no solution x_x

terenzreignz · Answer

Sure there are.

isaiah.feynman · Answer

Y=0 and Y=1

terenzreignz · Answer

That's right, those are the only two numbers which are their own square root.

Now, you have 
$$5-\frac{ 2 }{ x } = \sqrt{5-\frac{ 2 }{ x }}$$
Which technically reads "5 minus 2 over x is its own square root"

Having said that, what must 5 - 2/x be equal to?
$$\Large 5-\frac2x=\color{red}?$$

isaiah.feynman · Answer

X could be 0 or 1?

terenzreignz · Answer

x? no. because this $\large x = \sqrt x$ is not our equation.
What we have is 
$$5-\frac{ 2 }{ x } = \sqrt{5-\frac{ 2 }{ x }}$$

Which is, actually, just a fancier version of it, but the essence remains the same.

isaiah.feynman · Answer

I need a little push here. I just came across this question and thought it'd be interesting to solve.

terenzreignz · Answer

Right... here's a rather strong hint :P
$$\Large y = \sqrt y$$ implies y must either be 1 or 0.

Similarly, 
$$\Large z+1 = \sqrt{z+1}$$ implies z+1 must either be 1 or 0
or that z must either be -1 or 0.

Similarly still, 
$$\Large 3m +4 = \sqrt{3m+4}$$ implies 3m+4 must either be 1 or 0, or that
m must either be -4/3 or -1

Got it? XD

isaiah.feynman · Answer

YES!!!!

terenzreignz · Answer

so... applying that logic to your original question, what must x be?

isaiah.feynman · Answer

Lol x = 1/2, 2/5

terenzreignz · Answer

Brilliant ^_^
See? Not so hard :P

anonymous · Answer

y=0! Final answer.

terenzreignz · Answer

Actually, I think squaring both sides of the original equation actually complicates things XD

isaiah.feynman · Answer

Know what I did? I kept manipulating the exponents on both sides trying not to square.

terenzreignz · Answer

It took a little experience (knowledge that only 0 and 1 are their own squares/square roots)

isaiah.feynman · Answer

Oh wow, I have learned something new today in mathematics.