How do you evaluate $\sqrt{16 - x}$ $sqrt(16-x)$ when x = 8?

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How do you evaluate $\sqrt{16 - x}$ $sqrt(16-x)$ when x = 8?

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Answer 1 · 2018-03-10T12:23:29

$2 \sqrt{2}$ $2sqrt2$

Explanation:

First substitute 8 for x in the equation
$\sqrt{16 - 8}$ $sqrt(16-8$
Then simplify to get
$\sqrt{8}$ $sqrt 8$
This can be broken down into surd form to get
$\sqrt{4} . \sqrt{2}$ $sqrt4. sqrt2$
The square root of 4 is 2, but the square root of 2 is irrational, so can be left in surd form.
This leaves
$2 \sqrt{2}$ $2sqrt2$

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How do you evaluate $\sqrt{16 - x}$ $sqrt(16-x)$ when x = 8?

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How do you evaluate √16−xsqrt(16-x) when x = 8?

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Explanation:

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How do you evaluate $\sqrt{16 - x}$ $sqrt(16-x)$ when x = 8?