square root of 98 + cube root of 32 - sixth root of 50.

2. Originally Posted by varungk92
square root of 98 + cube root of 32 - sixth root of 50.
$\root {2}\of{98} \approx 9.8995$

$\root {3}\of{32} \approx 3.1748$

$\root {6}\of{50} \approx 1.9194$

so:

$
\root {2}\of{98} + \root {3}\of{32} +\root {6}\of{50} \approx 9.8995+3.1748+1.9194 \approx 14.994
$

RonL

3. Hello, varungk92!

Are you sure of the wording of the problem?
It doesn't simplify very much and it's a really stupid problem.

I suspect that it's suppose to be: . $\sqrt{98} + 3\sqrt{32} - 6\sqrt{50}$

You're expected to know how to simplify radicals: . $\sqrt{ab} \:=\:\sqrt{a}\cdot\sqrt{b}$

The first term is: . $\sqrt{98}\;=\;\sqrt{49\cdot2}\;=\;\sqrt{49}\cdot\s qrt{2} \;= \;7\sqrt{2}$

The second term is: . $3\sqrt{32}\;=\;3\sqrt{16\cdot2}\;=\;3\sqrt{16}\cdo t\sqrt{2}\;=\;3\cdot4\cdot\sqrt{2} \;=\;12\sqrt{2}$

The third term is: . $6\sqrt{50}\;=\;6\sqrt{25\cdot2}\;=\;6\cdot\sqrt{25 }\cdot\sqrt{2}\;=\;6\cdot5\cdot\sqrt{2}\;=\;30\sqr t{2}$

Then the problem becomes: . $7\sqrt{2} + 12\sqrt{2} - 30\sqrt{2}$

Answer: . $\boxed{-11\sqrt{2}}$