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Paraphin [41]
3 years ago
10

Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)

Mathematics
1 answer:
Anna71 [15]3 years ago
7 0
GIVEN:

5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)

remember:

\sqrt{u} = {u}^{ \frac{1}{2} }

And

{u}^{n} \times {u}^{m} = {u}^{n + m}

SOLVE:

start by multiplying the factors:

5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))

simplify by combing like terms. Most terms subtract off, leaving:

5( {u}^{6} - 1)

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

5( {u}^{3} - 1)( {u}^{3} + 1)

I'm not exactly sure what form they want the answer in...

so taking the square root:

\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }

so my best answer is:

{5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }
or the more factored form:

{5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!
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Step-by-step explanation:

<u><em>Step(I)</em></u>:-

The given function  h(t) = 3+70t - 16t²

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<em>Step(ii</em>):-

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