anonymous
  • anonymous
Medal + fan Simplify the expression. x times the square root of the quantity 8 x cubed, plus x times the square root of 2, plus four times the square root of the quantity 2x
Mathematics
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anonymous
  • anonymous
Medal + fan Simplify the expression. x times the square root of the quantity 8 x cubed, plus x times the square root of 2, plus four times the square root of the quantity 2x
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[x \sqrt{8x ^{3}} + x \sqrt{2} + 4\sqrt{2x}\]
anonymous
  • anonymous
I'm usually good at math but this question is a little confusing.... So far I think the answer is 2 because yo can simplify it to get 2x^2sqrt(2x)... and so on
anonymous
  • anonymous
\[2\sqrt{2} \times x \times x ^{3/2} + x \times \sqrt{2} + 4 \sqrt{2} \times x^{1/2} \]

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anonymous
  • anonymous
To write it into its raw exponential form.
anonymous
  • anonymous
why did you put fractions in the radical?
anonymous
  • anonymous
By putting the exponents in their fraction forms, you can clearly see their combined values.
anonymous
  • anonymous
This will allow you to factor out an exponential of x, through subtraction of the exponent.
anonymous
  • anonymous
As you can see in this question, the lowest "factorable" term is \[\sqrt{x}\]
anonymous
  • anonymous
I'm only in ninth grade, I don't think this is necessary.. besides, I haven't learned this yet.
anonymous
  • anonymous
I'm not looking for the lowest factorable term.... I'm looking for the leading coefficient when it's in simplified form
anonymous
  • anonymous
But it's factored form looks something like this: \[\sqrt{2} \sqrt{x} (4+\sqrt{x}+2 x^2)\]
anonymous
  • anonymous
To do that, you need write out the terms in full and subtract the exponents when factoring.
anonymous
  • anonymous
I think we're looking for the simplified form not the factored form... \[2x^2 \sqrt{2x} + x \sqrt{2} + 4 \sqrt{2x}\]
anonymous
  • anonymous
Is this fully simplified?
anonymous
  • anonymous
Well, I suppose you can't go any further to "simplify" it without factoring.
anonymous
  • anonymous
ok, so we agree that the leading coefficient is 2?
anonymous
  • anonymous
Okay, thanks, I got it right!
anonymous
  • anonymous
The leading coefficient of the entire term is not 2, its \[2\sqrt{2}\]
anonymous
  • anonymous
100% on my quiz! woot!
anonymous
  • anonymous
Great to hear!
anonymous
  • anonymous
Well, I guess it was the answer they were looking for!
anonymous
  • anonymous
As promised... medal and fan....

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