mutiplying radicals

Thread Starter

zelda1850

Joined Jan 3, 2010
18
am i doing these questions correctly can someone check

1) 3\(\sqrt{3}\)(4-3\(\sqrt{5}\))

3\(\sqrt{3}\) * (4) = 12\(\sqrt{3}\)
3\(\sqrt{3}\) * (3\(\sqrt{5}\) = 9\(\sqrt{15}\))

= 12\(\sqrt{3}\) - 9\(\sqrt{15}\)


2) 4\(\sqrt{15}\)(-3\(\sqrt{6}\) + 5)

4\(\sqrt{15}\) * (-3\(\sqrt{6}\) = 12\(\sqrt{90}\))
4\(\sqrt{15}\) * (5) = 20\(\sqrt{15}\)

12\(\sqrt{90}\) + 20\(\sqrt{15}\)


3) \(\sqrt{-2}\) (\(\sqrt{10}\) - 4\(\sqrt{6}\)

\(\sqrt{-2}\) * \(\sqrt{10}\) = \(\sqrt{-20}\)
\(\sqrt{-2}\) * 4\(\sqrt{6}\) = 4\(\sqrt{-12}\)

= \(\sqrt{-20}\) - 4\(\sqrt{-12}\)

4) (\(\sqrt{2a}\) - 5)(7\(\sqrt{2a}\) - 5)

-5 * -5 = -25
-5 * \(\sqrt{2a}\) = -10a

7\(\sqrt{2a}\) * -5 = -35 \(\sqrt{2a}\)
7\(\sqrt{2a}\) * 2a = 7\(\sqrt{2a}\)

and then i got stuck

5) (7+ \(\sqrt{6}\))(1 + \(\sqrt{6}\))

\(\sqrt{6}\) * \(\sqrt{6}\) = \(\sqrt{36}\)
\(\sqrt{6}\) * 7 = 7\(\sqrt{-12}\)

1 * \(\sqrt{6}\) = \(\sqrt{6}\)
1 * 7 = 7

= 7 + 7\(\sqrt{6}\) + \(\sqrt{36}\)
 
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