How can you factor out x from the original term?

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SUMMARY

The discussion clarifies how to factor out \(x\) from the expression \(\sqrt{2x^2 + 1}\). It establishes that for \(x > 0\), the expression can be rewritten as \(x\sqrt{2 + \frac{1}{x^2}}\). The key identity used is \(\sqrt{x^2} = x\), which is valid under the condition \(x > 0\). The general form of the equation is \(\sqrt{2x^2 + 1} = |x|\sqrt{2 + \frac{1}{x^2}}\) for \(x \neq 0\).

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In a text,

We have this:

$\sqrt{2x^2 + 1}$

is equal to

$x \sqrt{2 + \frac{1}{x}}$

I am confused as to how to factor out x from the original term.
 
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The two expressions are not equal. Actually,

$$\sqrt{2x^2 + 1} = x\sqrt{2 + \frac{1}{x^2}}\quad \text{for}\quad x > 0.$$

For if $x > 0$,

$$2x^2 + 1 = 2x^2 + \frac{x^2}{x^2} = x^2\Bigl(2 + \frac{1}{x^2}\Bigr),$$

so that

$$\sqrt{2x^2 + 1} = \sqrt{x^2\Bigl(2 + \frac{1}{x^2}\Bigr)} = \sqrt{x^2} \sqrt{2 + \frac{1}{x^2}} = x\sqrt{2 + \frac{1}{x^2}}.$$

Note the use of the identity $\sqrt{x^2} = x$, $x > 0$.

More generally,

$$\sqrt{2x^2 + 1} = |x|\sqrt{2 + \frac{1}{x^2}}\quad \text{for} \quad x \neq 0$$
 

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