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Question is in the title. My textbook has given me the question and solution but nothing in between. For this question I do not even know where to start.

In this chapter I've been learning about radicals, including the product and quotient rule, along with how to add and subtract radicals, rationalizing denominators and rational exponents.

Yet, for this particular question I'm struggling to even get started and understand what my first step would be?

I tried expressing the base as a exponent as opposed to a radical $50y^{2/8}$ to see if that would lead me down a path but it did not.

I find this question tricky because I am not sure which of the things I've been reading about in this chapter to apply here with a goal of simplifying the expression to $5y^4\sqrt{2}$?

Doug Fir
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    I suppose you mean $\sqrt{50y^8}$ in the title? – Yanko Jan 04 '19 at 16:36
  • It looks like you are missing a square in the second expression. $(5y^4\sqrt{2})^2$ gives you $25y^8\cdot 2$ – Chaos Jan 04 '19 at 16:37
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    Anyway, it looks like all you're missing the fact that for any two real numbers $a,b$ one has $\sqrt{ab} = \sqrt{a}\sqrt{b}$. So for instance since $50=25*2$ we have $\sqrt{50} = \sqrt{25}\sqrt{2} = 5\cdot \sqrt{2}$. – Yanko Jan 04 '19 at 16:39
  • Thanks for catching that, yes sorry and I've updated the title now too. – Doug Fir Jan 04 '19 at 16:39
  • Check this out https://math.stackexchange.com/questions/2047349/when-does-sqrta-b-sqrta-sqrtb – Yanko Jan 04 '19 at 16:40
  • Thanks @Yanko! I think I can take it from there following your feedback and comment. Should I delete? – Doug Fir Jan 04 '19 at 16:40
  • Someone answered so I'll accept when the timer comes off, thanks for your help here – Doug Fir Jan 04 '19 at 16:41
  • @DougFir You're welcome. – Yanko Jan 04 '19 at 16:42

1 Answers1

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Use that:

$$\sqrt{ab}=\sqrt a\sqrt b$$ (when $a,b>0$)

Particularly useful here is:

$$\sqrt{a^2b}=a\sqrt b$$

Rhys Hughes
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