Can someone help me with this problem and explain the steps I need help and I have a test for this type of question. I need to simplify $$-\sqrt{27k^7q^8}$$
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1$\begingroup$ Do you mean $-\sqrt{27k^7q^8}$? $\endgroup$– homegrownCommented May 3, 2014 at 21:18
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$\begingroup$ yes please help? $\endgroup$– CrissyCommented May 3, 2014 at 21:22
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1 Answer
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Anything which is a square you can take out of the radical. In particular, $27k^7q^8$ contains a lot of squares:
$27 = 9 \cdot 3$ ($9$ is a square--it is the square of $3$)
$k^7 = k^6 \cdot k$ ($k^6$ is a square--it is the square of $k^3$)
$q^8$ is a square (it is the square of $q^4$)
So you can write $$ 27 k^7 q^8 = (3 \cdot k^3 \cdot q^4)^2 \cdot 3 \cdot k $$ Then $$ -\sqrt{27k^7q^8} = -\sqrt{(3 \cdot k^3 \cdot q^4)^2 \cdot 3 \cdot k} = -3 k^3 q^4 \sqrt{3k}. $$
answered May 3, 2014 at 21:30
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$\begingroup$ Thank you for the explanation. But can you explain why K^6 is a square of k^3 $\endgroup$– CrissyCommented May 3, 2014 at 21:42
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$\begingroup$ @Crissy The square of $k$ is $k \cdot k$, the square of $k \cdot k$ is $k \cdot k \cdot k \cdot k$, and so on. If you square $k^3$, you double the number of $k$s. So you now have $6$ $k$s, which is $k^6$. $\endgroup$ Commented May 3, 2014 at 22:41