I need help, I know how to do it but i'm not getting the right answer The square root of 5 minus the square root of two over four square root of two
\[\sqrt{5}-\sqrt{2}/4\sqrt{2}\]
\[\large \sqrt{5} - \sqrt{\frac{ 2 }{ 4\sqrt{2} }}\]
this is the answer?
it is not easy to tell what the thing means, yours or mine, i read it as .. "square root of" ( that whole quantity)
i posted what i meant on the under it
probably yours is it, you can cancel the root 2 in the second, root(5) - 1/4 thats it
well the problem isn't actually a word problem i just didn't know how to type it, but can you explain
\[\large \sqrt{5} - \sqrt{\frac{ 2 }{ 4\sqrt{2} }}\] \[\large \sqrt{5} - \frac{ \sqrt{2} }{ \sqrt{(4*\sqrt{2}}) }\]
cause i thought you had to do 5√−2√/4√2 * 4√2/4√2
to cancel out the radical in the denominator
\[\large \sqrt{5} - \frac{ \sqrt{2} }{ \sqrt{4}*2 }\] \[\sqrt{5}-\frac{ \sqrt{2} }{ 4 }\]
that is the simplified version of the prob i started with... oh ok, then you can cancel the root 2's like before, and then
\[\sqrt{5}-\frac{ \sqrt{2} }{ 4\sqrt{2} } = \sqrt{5} - \frac{ 1 }{ 4 }\]
maybe think of the second term as individual terms or fractions \[\frac{ 1 }{ 4 }*\frac{ \sqrt{2} }{ \sqrt{2} }\]
ok can you help me with one more problem
sure
\[5\sqrt{6}/\sqrt{15}\]
using the exponent rules, \[\large \frac{ 5\sqrt{6} }{ \sqrt{15} } = \frac{ 5*\sqrt{3*2} }{ \sqrt{5*3} }=\frac{ 5\sqrt{3} \sqrt{2} }{ \sqrt{5} \sqrt{3}}\] the root3 cancels in top and bottom \[\frac{ 5\sqrt{2} }{ \sqrt{5} }\]
most the time they dont want a root on the denominator, so then multiply by root5/root5 \[\frac{ 5\sqrt{2} }{ \sqrt{5} } * \frac{ \sqrt{5} }{ \sqrt{5} } = \frac{ 5\sqrt{2}\sqrt{5} }{ 5 }\]
that can be simplified more even, cancel the 5/5, and the leftover is just root2 times root 5 \[\sqrt{2}\sqrt{5} = \sqrt{10}\] simplifies to square root of 10
ok thank you
welcome
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