# Thread: Radicals #7

1. ## Radicals #7

Question #3 a)b)c)d): imgur: the simple image sharer

My attempt: imgur: the simple image sharer

I have no clue how I should start c) or d), I tried a) and b) but got it wrong...

2. ## Re: Radicals #7

Hello, Mathnood768!

Simplify. Identify the variable for which the radical represents a real number.

$3(a)\;\;\text{-}3\left(\sqrt{2} - 4\right) + 9\sqrt{2}$
We have: . $\text{-}3\sqrt{2} + 12 + 9\sqrt{2} \;=\;12 + 6\sqrt{2}$

$(b)\;\;7\left(\text{-}1 - 2\sqrt{6}\right) + 5\sqrt{6} + 8$
We have: . $-7 - 14\sqrt{6} + 5\sqrt{6} + 8 \;=\;1 - 9\sqrt{6}$

$(c)\;\;4\sqrt{5}\left(\sqrt{3j} + 8\right) - 3\sqrt{15j} + \sqrt{5}$
We have: . $4\sqrt{15j} + 32\sqrt{5} - 3\sqrt{15j} + \sqrt{5} \;=\;\sqrt{15j} + 33\sqrt{5}$

If $\sqrt{15j}$ is real, then $j \ge 0$

$(d)\;\;3 - \sqrt[3]{4k}\left(12 + 2\sqrt[3]{8}\right)$

We have: . $3 - \sqrt[3]{4k}(12 + 2\!\cdot\!2) \;=\;3 - \sqrt[3]{4k}(12 + 4) \;=\;3 - 16\sqrt[3]{4k}$

If $\sqrt[3]{4k}$ is real, then $k$ = any real number.

Thank you!