Simplifying a square root expression

[tmt1]

I have this expression:

$$\sqrt{ 1 - \frac{16}{\sqrt{x^2 + 16}}}$$

And the textbook simplifies it to

$$\frac{x}{\sqrt{x^2 + 16}}$$

But I'm not sure how it does this.

 

[MarkFL]

I can see the following being true:

$$\sqrt{1-\frac{16}{x^2+16}}=\frac{|x|}{\sqrt{x^2+16}}$$

But what you say your textbook is implying is not true.

 

[Wilmer]

I have this expression:

$$\sqrt{ 1 - \frac{16}{\sqrt{x^2 + 16}}}$$

And the textbook simplifies it to

$$\frac{x}{\sqrt{x^2 + 16}}$$

But I'm not sure how it does this.

Suggestions (in general):
1: during simplification process, let k = SQRT(x^2 + 16);
saves significant "time"

2: substitute a value for x: then check if original
expression = book's expression

 

[Deveno]

I have this expression:

$$\sqrt{ 1 - \frac{16}{\sqrt{x^2 + 16}}}$$

And the textbook simplifies it to

$$\frac{x}{\sqrt{x^2 + 16}}$$

But I'm not sure how it does this.

Write:

$\sqrt{ 1 - \dfrac{16}{\sqrt{x^2 + 16}}} = \sqrt{\dfrac{x^2 + 16}{x^2 + 16} - \dfrac{16}{x^2 + 16}}$

and simplify the numerator under the radical.

As MarkFL implies, unless the domain of $x$ is known to be non-negative, we have:

$\sqrt{x^2} = |x|$, not $x$.

 

[mrtwhs]

Write:

$\sqrt{ 1 - \dfrac{16}{\sqrt{x^2 + 16}}} = \sqrt{\dfrac{x^2 + 16}{x^2 + 16} - \dfrac{16}{x^2 + 16}}$

and simplify the numerator under the radical.

As MarkFL implies, unless the domain of $x$ is known to be non-negative, we have:

$\sqrt{x^2} = |x|$, not $x$.

$\sqrt{ 1 - \dfrac{16}{\sqrt{x^2 + 16}}} = \sqrt{\dfrac{x^2 + 16}{x^2 + 16} - \dfrac{16}{x^2 + 16}}$

but this is clearly not true.

 

[Wilmer]

but this is clearly not true.

Looks to me like a ye olde typo, nuttin' else !

 

[Deveno]

$\sqrt{ 1 - \dfrac{16}{\sqrt{x^2 + 16}}} = \sqrt{\dfrac{x^2 + 16}{x^2 + 16} - \dfrac{16}{x^2 + 16}}$

but this is clearly not true.

Indeed, I missed an extra radical in the denominator, leaving me to wonder what the text in question actually says...