MHB Simplifying a square root expression

AI Thread Summary
The discussion centers on the simplification of the expression √(1 - 16/√(x² + 16)), which a textbook claims simplifies to x/√(x² + 16). Participants express confusion over this simplification, noting that the correct approach involves rewriting the expression to clarify the numerator. It is highlighted that unless x is non-negative, the simplification should account for the absolute value of x. The conversation suggests that a potential error in the textbook may exist, leading to the conclusion that the simplification provided is not accurate. The overall consensus is that careful attention to the domain and proper algebraic manipulation is essential for accurate simplification.
tmt1
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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.
 
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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.
 
tmt said:
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
 
tmt said:
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 $x$.
 
Deveno said:
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.
 
mrtwhs said:
but this is clearly not true.
Looks to me like a ye olde typo, nuttin' else !
 
mrtwhs said:
$\sqrt{ 1 - \dfrac{16}{\sqrt{x^2 + 16}}} = \sqrt{\dfrac{x^2 + 16}{x^2 + 16} - \dfrac{16}{x^2 + 16}}$

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...
 
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