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anonymous

  • 4 years ago

The square root of 8x times the square root of 8x

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  1. nikita2
    • 4 years ago
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    \[\sqrt{8x}\sqrt{8x}\] ?

  2. nikita2
    • 4 years ago
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    = 8x and x>=0

  3. anonymous
    • 4 years ago
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    Can you explain better how you got that?

  4. TuringTest
    • 4 years ago
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    \[\sqrt a\times\sqrt a=(\sqrt a)^2=a\]

  5. anonymous
    • 4 years ago
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    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

  6. anonymous
    • 4 years ago
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    But i understand how the square roots cancel out

  7. nikita2
    • 4 years ago
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    \[\sqrt{5}/\sqrt{8x} = \sqrt{\left(\begin{matrix}5 \\ 8x\end{matrix}\right)}\]

  8. anonymous
    • 4 years ago
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    I need to rationalize the denominator

  9. TuringTest
    • 4 years ago
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    \[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{2^3x}}\cdot\frac{\sqrt{2x}}{\sqrt{2x}}=\frac{\sqrt{10x}}{4x}\]

  10. anonymous
    • 4 years ago
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    Why don't i multiply the numerator and denominator by the swuare root of 8x?? why do i have to break it up

  11. TuringTest
    • 4 years ago
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    \[\frac{\sqrt5}{\sqrt{8x}}=\frac{\sqrt5}{\sqrt{8x}}\cdot\frac{\sqrt{8x}}{\sqrt{8x}}=\frac{\sqrt{40x}}{\sqrt{64x^2}}=\frac{2\sqrt{10x}}{8x}=\frac{\sqrt{10x}}{4x}\]

  12. anonymous
    • 4 years ago
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    wait i thought you said \[\sqrt{8x } * \sqrt{8x} = 8x\]

  13. TuringTest
    • 4 years ago
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    yeah it does, look at the denominator. it winds up as 8x, then simplifies to 4x

  14. anonymous
    • 4 years ago
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    and on the last step you just divided the top and bottum by 2 to get rid of it?

  15. TuringTest
    • 4 years ago
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    yeah

  16. anonymous
    • 4 years ago
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    ok thank you!

  17. TuringTest
    • 4 years ago
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    no prob

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