Can you show this?

Question

Show that $$\frac{\sqrt{2x^2+1}}{x}=\sqrt{2+1/x^2}.$$

This seems like basic algebra, but I can't really manage to deal with the sqrt sign and prove this.

Answer 1

I think it's wrong.

Try $x=-1$.

By the way, $$\frac{\sqrt{2x^2+1}}{x}=\sqrt{2+\frac{1}{x^2}}$$ for $x>0$ and

$$\frac{\sqrt{2x^2+1}}{x}=-\sqrt{2+\frac{1}{x^2}}$$ for $x<0$.

The both equalities we can write so: $$\sqrt{2+\frac{1}{x^2}}=\frac{\sqrt{2x^2+1}}{\sqrt{x^2}}=\frac{\sqrt{2x^2+1}}{|x|}$$

Answer 2

1) $x>0,$ the equality is OK.

2) $x < 0$, LHS is negative, RHS positive.

LHS: $\dfrac{\sqrt{2x^2 +1} }{x}=$

$\dfrac {\sqrt{x^2(2+\dfrac{1}{x^2})}}{x} =$

$\dfrac{|x|\sqrt{2+\dfrac{1}{x^2}}}{x} =$

$\dfrac{|x|}{x}\sqrt{2+\dfrac{1}{x^2}}.$

Note:

$\dfrac{|x|}{x} = 1$ for $x>0;$

$\dfrac{|x|}{x} = -1$ for $x<0.$