Question #bfe87

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Question #bfe87

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Answer 1 · 2016-09-19T11:07:41

$\pm 2 \sqrt{3}$ $+-2sqrt(3)$

Explanation:

You are looking for squared numbers that you can 'remove' from the square root

Known: $3 \times 4 = 12$ $" "3xx4=12$

But 4 is the same as $2^{2}$ $2^2$ so we can write 12 as $3 \times 2^{2}$ $3xx2^2$ giving:

$\sqrt{3 \times 2^{2}}$ $sqrt(3xx2^2)$

Take the $2^{2}$ $2^2$ outside the square root giving:

$2 \sqrt{3} \leftarrow part answer$ $2sqrt(3)" "color(brown)( larr" part answer")$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Observe that in taking the $2^{2}$ $2^2$ out of the square root we have to apply $\sqrt{2^{2}} = 2$ $sqrt(2^2)=2$

Actually it is $\pm 2$ $+-2$ as : $(- 2) \times (- 2) = (+ 2) \times (+ 2) = 4$ $(-2)xx(-2)=(+2)xx(+2)=4$

So if you wish to be totally correct you would write:

$\sqrt{12} = \pm 2 \sqrt{3}$ $sqrt(12)=+-2sqrt(3)$

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Question #bfe87

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