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kirk.freedman

Simplify (sqrt x + sqrt 3) (sqrt x + sqrt 27) show work or explain please

  • one year ago
  • one year ago

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  1. kirk.freedman
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    • one year ago
  2. sauravshakya
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    Hint: (a+b) (c+d) = ac +ad + bc +bd

    • one year ago
  3. kirk.freedman
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    do i keep the roots?

    • one year ago
  4. sauravshakya
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    yes first keep the roots.... tell me what u got

    • one year ago
  5. kirk.freedman
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    ok so I got \[\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{27}+\sqrt{3}\sqrt{x}+\sqrt{3}\sqrt{27}\]

    • one year ago
  6. sauravshakya
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    good

    • one year ago
  7. sauravshakya
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    |dw:1356401518093:dw|Now,

    • one year ago
  8. sauravshakya
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    Can u use it too

    • one year ago
  9. kirk.freedman
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    \[x+\sqrt{27x}+\sqrt{3x}+9\]

    • one year ago
  10. kirk.freedman
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    or i guess\[x+\sqrt{30x}+9\]

    • one year ago
  11. sauravshakya
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    good

    • one year ago
  12. kirk.freedman
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    Now what??

    • one year ago
  13. kirk.freedman
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    I figured out the answer is \[x+4\sqrt{3x}+9\] But I do not understand how you get \[4\sqrt{3x}\] from \[\sqrt{27x}+\sqrt{3x}\]

    • one year ago
  14. calculusfunctions
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    @kirk.freedman \[\sqrt{27x}+\sqrt{3x}\neq \sqrt{30x}\]Therefore your above answer is incorrect. I don't know why @sauravshakya said "good"? I know he knows better than that. Maybe he was tired or rushing. You can only add or subtract two radicals by adding or subtracting the numbers in front iff they have the same index and the same radicand. You can multiply or divide two radicals by multiplying or dividing their radicands iff they have the same index. The only thing wrong about your answer is the root 30x. you should simplify the sq.root of 27x first. Like this:\[\sqrt{27x}=\sqrt{9}\sqrt{3x}=3\sqrt{3x}\] Thus\[(\sqrt{x}+\sqrt{3})(\sqrt{x}+\sqrt{27})\]\[=x +\sqrt{27x}+\sqrt{3x}+\sqrt{81}\]\[=x +3\sqrt{3x}+\sqrt{3x}+9\]\[=x +4\sqrt{3x}+9\]And this @kirk.freedman is your final answer!

    • one year ago
  15. kirk.freedman
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    So if \[\sqrt{27x}= 3\sqrt{3}\] you still have the other \[3\sqrt{3}\] don't you?

    • one year ago
  16. kirk.freedman
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    I'm sorry, add an x into those above.

    • one year ago
  17. calculusfunctions
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    No the other one is just root 3. 3 root 3 is only derived from simplifying root 27. Understand?

    • one year ago
  18. calculusfunctions
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    Well x or no x, 3 root 3 is only derived from simplifying root 27.

    • one year ago
  19. kirk.freedman
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    oh wait so rt27x= 3rt3x, and still have the rt3x though??

    • one year ago
  20. calculusfunctions
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    Yes!

    • one year ago
  21. kirk.freedman
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    from the equation above \[x+\sqrt{27x}+\sqrt{3x}+9\] See we still have the \[\sqrt{3}\]?? do we just get rid of it or what? Because we end up having \[3\sqrt{3x}+\sqrt{3x}\]

    • one year ago
  22. calculusfunctions
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    Did you not read every single line of my explanation above, because I explained this very clearly.

    • one year ago
  23. kirk.freedman
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    OHHHHHHHHH I got it, thanks so much!

    • one year ago
  24. calculusfunctions
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    3 root 3 + 1 root 3 = 4 root 3 because 3 + 1 = 4.

    • one year ago
  25. calculusfunctions
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    Good, Excellent!

    • one year ago
  26. sauravshakya
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    Sorry I was saying good for the first one... And I was unable to continue as my internet stop working

    • one year ago
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