How do you simplify $\frac{\sqrt{h^{3}}}{\sqrt{8}}$ $sqrt(h^3)/sqrt8$ ?

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How do you simplify $\frac{\sqrt{h^{3}}}{\sqrt{8}}$ $sqrt(h^3)/sqrt8$ ?

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Answer 1 · 2017-12-14T12:19:12

$\pm \frac{h \sqrt{2 h}}{4}$ $+-(hsqrt(2h))/4$

Explanation:

There are two things to consider.

1) Are there any squared values we can 'take out' of the roots?

2) It is considered not good practice to have a root as or in the denominator so we need to change what is left of it into a whole number.

Given: $\frac{\sqrt{h^{3}}}{\sqrt{8}}$ $sqrt(h^3)/sqrt(8)$

Write as: $\frac{\sqrt{h^{2} \times h}}{\sqrt{2^{2} \times 2}} ddd giving: ddd \frac{h \sqrt{h}}{2 \sqrt{2}}$ $sqrt(h^2xxh)/(sqrt(2^2xx2))color(white)("ddd")" giving: "color(white)("ddd") (hsqrt(h))/(2sqrt(2))$

Now we 'get rid of the root in the denominator; multiply by 1 and you do not change the inherent value. However 1 comes in many forms.

$\frac{h \sqrt{h}}{2 \sqrt{2}} \times 1 dddd \to dddd \frac{h \sqrt{h}}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $color(green)((hsqrt(h))/(2sqrt(2))color(red)(xx1) color(white)("dddd")-> color(white)("dddd")(hsqrth)/(2sqrt2)color(red)(xxsqrt2/sqrt2) )$

$d ddddddddddd \to dddd \frac{h \sqrt{2 h}}{4}$ $color(white)("d")color(green)(color(white)("ddddddddddd")->color(white)("dddd")(hsqrt(2h)) /4 )$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Foot note$ $color(blue)("Foot note")$

Compare the consequence of $\sqrt{h} \times \sqrt{2} \to \sqrt{2 h} d$ $sqrt(h)xxsqrt(2)->sqrt(2h)color(white)("d")$ to the example:

$\sqrt{4} \times \sqrt{9} d = d 2 \times 3 = 6$ $sqrt(4)xxsqrt(9)color(white)("d")=color(white)("d")2xx3=6$

$\sqrt{4 \times 9} d = d \sqrt{36} d = d 6$ $sqrt(4xx9)color(white)("d")=color(white)("d")sqrt(36)color(white)("d")=color(white)("d")6$

$So h \sqrt{h} \times \sqrt{2} dd = dd h \sqrt{h \times 2} dd = dd h \sqrt{2 h}$ $color(brown)("So " hsqrth xxsqrt2color(white)("dd") =color(white)("dd") hsqrt(hxx2)color(white)("dd") =color(white)("dd") hsqrt(2h))$

$d$ $color(white)("d")$

$and 2 \sqrt{2} \times \sqrt{2} d = d 2 \sqrt{2 \times 2} d = d 2 \sqrt{4} d = d 2 \times 2 = 4$ $color(brown)("and "2sqrt(2) xxsqrt2color(white)("d") =color(white)("d") 2sqrt(2xx2)color(white)("d") =color(white)("d") 2sqrt(4)color(white)("d") =color(white)("d") 2xx2=4$

Technically the answer should be $\pm \frac{h \sqrt{2 h}}{4}$ $+-(hsqrt(2h))/4$

As the square root of a number is $\pm$ $+-$

Example:

$(- 2) \times (- 2) d = d (+ 4) d = d (+ 2) \times (+ 2)$ $(-2)xx(-2)color(white)("d")=color(white)("d")(+4)color(white)("d")=color(white)("d")(+2)xx(+2)$

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