How do you simplify ${\sqrt{16 b}}^{4}$ $sqrt(16b)^4$ ?

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Answer 1 · 2015-04-07T04:43:01

$\sqrt{16 \cdot b^{4}}$ $sqrt(16*b^4)$ = $\pm 4 b^{2}$ $+-4b^2$
We use the Identity $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ $sqrt(a*b) = sqrta*sqrtb$

$\sqrt{16 \cdot b^{4}}$ $sqrt(16*b^4)$

= $\sqrt{4^{2} \cdot b^{4}}$ $sqrt(4^2*b^4)$

= $\sqrt{4^{2}} \cdot \sqrt{b^{4}}$ $sqrt(4^2)*sqrt(b^4)$

= $4 \cdot b^{2}$ $4*b^2$

Squaring $4 \cdot b^{2}$ $4*b^2$ will give us $16 \cdot b^{4}$ $16*b^4$ . And squaring $- (4 \cdot b^{2})$ $-(4*b^2)$ will also give us $16 \cdot b^{4}$ $16*b^4$

Hence $\sqrt{16 \cdot b^{4}}$ $sqrt(16*b^4)$ = $\pm 4 b^{2}$ $+-4b^2$

How do you simplify sqrt(16b)^4√16b4sqrt(16b)^4?