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anonymous

  • one year ago

I need help with simplifying

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  1. anonymous
    • one year ago
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    \[6a \sqrt{80b}-\sqrt{180a ^{2}b}+5a \sqrt{b}\]

  2. Jhannybean
    • one year ago
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    Look at the middle term and take the prime factorization of 180

  3. Jhannybean
    • one year ago
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    Do the same for 80

  4. anonymous
    • one year ago
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    60 & 2 for 180 then 40&2 for 80 right?

  5. Jhannybean
    • one year ago
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    \[6a\sqrt{20\cdot 2 \cdot 2 \cdot b}- \sqrt{20 \cdot 9 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]

  6. Jhannybean
    • one year ago
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    \[60 \cdot 2 \ne 180\]

  7. anonymous
    • one year ago
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    lol i had meant 90

  8. Jhannybean
    • one year ago
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    Yes that works too :)

  9. Jhannybean
    • one year ago
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    now factor both 20's and the 9 into simpler, prime factors.

  10. anonymous
    • one year ago
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    10 x 2 for 20's & 3 x 3 for 9

  11. Jhannybean
    • one year ago
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    Good. \[6a\sqrt{10\cdot 2\cdot 2 \cdot 2 \cdot b}- \sqrt{10 \cdot 2 \cdot 3\cdot 3 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]

  12. Jhannybean
    • one year ago
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    factor both 10's one more time into prime factors.

  13. anonymous
    • one year ago
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    5 & 2

  14. Jhannybean
    • one year ago
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    Great \(\checkmark\) \[6a\sqrt{5\cdot 2\cdot 2\cdot 2 \cdot 2 \cdot b}- \sqrt{5\cdot 2 \cdot 2 \cdot 3\cdot 3 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]

  15. Jhannybean
    • one year ago
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    Now for every fair of 2 numbers, 1 of them will be extracted from the square root and multiplied to the term outside the square root.

  16. Jhannybean
    • one year ago
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    pair*

  17. Jhannybean
    • one year ago
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    For example, to start you off. \[6a\sqrt{5\color{blue}{\cdot 2\cdot 2}\color{red}{\cdot 2 \cdot 2} \cdot b} = 6a \cdot \color{blue}{2} \cdot \color{red}{2}\sqrt{5 \cdot b}\]

  18. anonymous
    • one year ago
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    \[48a \sqrt{5\times2timesb}-6a \sqrt{5\times2\times3 \times a \times b}+5a \sqrt{b}\]

  19. Jhannybean
    • one year ago
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    Not quite. In the first part you still have a 2 left within the square root when its been paired off and pulled out already. You should just have a 5 and a b left within the square root there since they are lone pairs and cannot be pulled out. Once things are paired off, they are not rewritten within the square root. Second part, the 6a looks right, therefore the only thing that should be left within the square root should be 5 and b.

  20. anonymous
    • one year ago
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    ohh okay i think i got it now

  21. Jhannybean
    • one year ago
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    It should look like this: \[6a\sqrt{5\color{blue}{\cdot 2\cdot 2}\color{red}{\cdot 2 \cdot 2} \cdot b} = 6a \cdot \color{blue}{2} \cdot \color{red}{2}\sqrt{5 \cdot b} = 24a\sqrt{5b}\]\[- \sqrt{5\cdot\color{red}{ 2 \cdot 2} \cdot \color{blue}{3\cdot 3} \cdot \color{green}{a\cdot a} \cdot b} = -6a\sqrt{5b}\]

  22. anonymous
    • one year ago
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    yeah thats what i got

  23. anonymous
    • one year ago
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    |dw:1441173348642:dw|

  24. Jhannybean
    • one year ago
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    \[(24a-6a)\sqrt{5b} +5a\sqrt{b}= 18a\sqrt{5b}+5a\sqrt{b} \]

  25. Jhannybean
    • one year ago
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    Coool, good job!

  26. anonymous
    • one year ago
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    yup, and thank you for your help

  27. Jhannybean
    • one year ago
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    Np.

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