How do you simplify $\sqrt{w^{14}}$ $sqrt(w^14)$ ?

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How do you simplify $\sqrt{w^{14}}$ $sqrt(w^14)$ ?

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Answer 1 · 2015-09-30T10:44:06

$= w^{7}$ $= color(blue)(w^7$

Explanation:

$\sqrt{w^{14}}$ $sqrt(w^14)$

Square root , can also be called as second root or a half power ( $\frac{1}{2}$ $color(blue)(1/2$ )

so , $\sqrt{w^{14}} = {(w^{14})}^{\frac{1}{2}}$ $sqrt (w^14) = (w^14)^(color(blue)(1/2)$

$= w^{7}$ $= color(blue)(w^7$

Answer 2 · 2015-10-01T22:31:07

$\sqrt{w^{14}} = ∣ ∣ w^{7} ∣ ∣$ $sqrt(w^14) = abs(w^7)$

Explanation:

If $a \geq 0$ $a >= 0$ then $\sqrt{a^{2}} = a$ $sqrt(a^2) = a$

If $a \geq 0$ $a >= 0$ and $b, c > 0$ $b, c > 0$ then ${(a^{b})}^{c} = a^{b c}$ $(a^b)^c = a^(bc)$

So if $w \geq 0$ $w >= 0$ then:

$\sqrt{w^{14}} = \sqrt{{(w^{7})}^{2}} = w^{7}$ $sqrt(w^14) = sqrt((w^7)^2) = w^7$

and if $w < 0$ $w < 0$ then:

$\sqrt{w^{14}} = \sqrt{{(- w)}^{14}} = \sqrt{{({(- w)}^{7})}^{2}} = {(- w)}^{7} = - w^{7}$ $sqrt(w^14) = sqrt((-w)^14) = sqrt(((-w)^7)^2) = (-w)^7 = -w^7$

So in both cases we have

$\sqrt{w^{14}} = ∣ ∣ w^{7} ∣ ∣$ $sqrt(w^14) = abs(w^7)$

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How do you simplify $\sqrt{w^{14}}$ $sqrt(w^14)$ ?

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