anonymous
  • anonymous
The square root of 8x times the square root of 8x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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nikita2
  • nikita2
\[\sqrt{8x}\sqrt{8x}\] ?
nikita2
  • nikita2
= 8x and x>=0
anonymous
  • anonymous
Can you explain better how you got that?

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TuringTest
  • TuringTest
\[\sqrt a\times\sqrt a=(\sqrt a)^2=a\]
anonymous
  • anonymous
Actually the original problem is the square root of 5 divided by the square root of 8x. Can you show the whole process because I'm really confused
anonymous
  • anonymous
But i understand how the square roots cancel out
nikita2
  • nikita2
\[\sqrt{5}/\sqrt{8x} = \sqrt{\left(\begin{matrix}5 \\ 8x\end{matrix}\right)}\]
anonymous
  • anonymous
I need to rationalize the denominator
TuringTest
  • TuringTest
\[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{2^3x}}\cdot\frac{\sqrt{2x}}{\sqrt{2x}}=\frac{\sqrt{10x}}{4x}\]
anonymous
  • anonymous
Why don't i multiply the numerator and denominator by the swuare root of 8x?? why do i have to break it up
TuringTest
  • TuringTest
\[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{8x}}\cdot\frac{\sqrt{8x}}{\sqrt{8x}}=\frac{\sqrt{40x}}{\sqrt{64x^2}}=\frac{2\sqrt{10x}}{8x}=\frac{\sqrt{10x}}{4x}\]
anonymous
  • anonymous
wait i thought you said \[\sqrt{8x } * \sqrt{8x} = 8x\]
TuringTest
  • TuringTest
yeah it does, look at the denominator. it winds up as 8x, then simplifies to 4x
anonymous
  • anonymous
and on the last step you just divided the top and bottum by 2 to get rid of it?
TuringTest
  • TuringTest
yeah
anonymous
  • anonymous
ok thank you!
TuringTest
  • TuringTest
no prob

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