# How do you simplify 4 sqrt 3 - sqrt 64 + 6 sqrt 27?

Nov 7, 2015

22sqrt3−8

#### Explanation:

We know that $\sqrt{64} = 8$, so we have

4sqrt3−8+6sqrt27

Now let's rewrite $\sqrt{27}$ as $\sqrt{9} \cdot \sqrt{3}$
since we know the square root of 9, we can write it as $3 \sqrt{3}$

Now just a bit of simple multiplication and combining link terms
4sqrt3−8+6(3sqrt3)

4sqrt3−8+18sqrt3)

22sqrt3−8

Nov 7, 2015

22sqrt[3] -8

#### Explanation:

We can simplify first by solving sqrt[64] since we know it is a perfect square. This leaves:

4sqrt[3] -8 + 6sqrt[27]

Then we need to see if we can simplify the square roots.

4sqrt[3] is as simplified as it gets.
6sqrt[27] can be broken up into factors so:

6sqrt[27] = 6sqrt[3*9]

9 is a perfect square so it can be rooted and brought out side of the square root. This leaves:

6(3)sqrt[3]

So now we have:

4sqrt[3] -8 + 18sqrt[3]

Combine terms:

22sqrt[3] - 8
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}