How do you simplify $8 \sqrt{11 b^{4}} - \sqrt{99 b^{4}}$ $8sqrt(11b^4) - sqrt(99b^4)$ ?

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Answer 1 · 2015-05-18T13:40:50

By exponential definition, we know that $\sqrt[n]{a^{m}} = a^{\frac{m}{n}}$ $root(n)(a^m)=a^(m/n)$

In this case, we can do the same for $b$ $b$ , factoring it out:

$8 b^{2} \sqrt{11} - b^{2} \sqrt{99}$ $8b^2sqrt(11)-b^2sqrt(99)$
However, we can see that $99 = 11 \cdot 9$ $99=11*9$ , so

$8 b^{2} \sqrt{11} - b^{2} \sqrt{11 \cdot 9}$ $8b^2sqrt(11)-b^2sqrt(11*9)$

However, $9$ $9$ is a square number. Thus, we can take its root out of the square root.

$8 b^{2} \sqrt{11} - 3 b^{2} \sqrt{11}$ $8b^2sqrt(11)-3b^2sqrt(11)$

However, as both factors are multiplying $b^{2} \sqrt{11}$ $b^2sqrt(11)$ , we can factor it in function of such elements:

$b^{2} \sqrt{11} (8 - 3)$ $b^2sqrt(11)(8-3)$

How do you simplify 8√11b4−√99b48sqrt(11b^4) - sqrt(99b^4)?