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anonymous

  • one year ago

simplify 5 square root of 7 + 12 square root of 6 - 10 square root of 7 - 5 square root of 6.

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  1. anonymous
    • one year ago
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    i need help please

  2. Michele_Laino
    • one year ago
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    hint: we can factor out square root of 7 between the first and the third term, furthermore, we can factor out square root of 6 between the second and fourth term, so we can write the subsequent step: \[\Large \begin{gathered} 5\sqrt 7 + 12\sqrt 6 - 10\sqrt 7 - 5\sqrt 6 = \hfill \\ \hfill \\ = \sqrt 7 \left( {5 - 10} \right) + \sqrt 6 \left( {12 - 5} \right) = ... \hfill \\ \end{gathered} \]

  3. Michele_Laino
    • one year ago
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    please complete

  4. anonymous
    • one year ago
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    honestly i dont know mate im really confused sorry... @Michele_Laino

  5. Michele_Laino
    • one year ago
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    following the rules of algebra of radicals, you can only sum similar radicals, namely radicals which have the same square roots as in your case

  6. Michele_Laino
    • one year ago
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    now, \[5\sqrt 7 ,\; - 10\sqrt 7 \] are similar since they both have square root of 7, right?

  7. anonymous
    • one year ago
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    i was trying to work it out and the answer i got was 5 square root of 7 - 7 quare root of 6

  8. anonymous
    • one year ago
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    @Michele_Laino

  9. Michele_Laino
    • one year ago
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    I got this since 5-10= -5, we have: \[\sqrt 7 \left( {5 - 10} \right) = - 5\sqrt 7 \]

  10. Michele_Laino
    • one year ago
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    furthermore 12-5=7, so we can write: \[\sqrt 6 \left( {12 - 5} \right)\]

  11. Michele_Laino
    • one year ago
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    oops.. \[\sqrt 6 \left( {12 - 5} \right) = 7\sqrt 6 \]

  12. anonymous
    • one year ago
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    and that will be 7 square root of 6 - 5 square root of 7 @Michele_Laino

  13. Michele_Laino
    • one year ago
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    that's right!

  14. anonymous
    • one year ago
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    thank you so much can you help me with one more? @Michele_Laino

  15. Michele_Laino
    • one year ago
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    ok!

  16. anonymous
    • one year ago
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    simplify square root of 5 (10-4 square root of 2) @Michele_Laino

  17. Michele_Laino
    • one year ago
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    we have to apply the distributive property of multiplication over addition, so we can write this: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt 5 \sqrt 2 \]

  18. Michele_Laino
    • one year ago
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    am I right?

  19. anonymous
    • one year ago
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    @Michele_Laino is that the answer to the question?

  20. Michele_Laino
    • one year ago
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    no, we have to write another step

  21. anonymous
    • one year ago
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    ohhhhhh okay well as i simplify it i got 5 square root of 2 - 4 square root of 10 @Michele_Laino

  22. Michele_Laino
    • one year ago
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    more precisely we can write this: \[\sqrt 5 \sqrt 2 = \sqrt {5 \times 2} = \sqrt {10} \]

  23. anonymous
    • one year ago
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    yea thats what i got @Michele_Laino

  24. Michele_Laino
    • one year ago
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    so we get: \[\sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} \]

  25. Michele_Laino
    • one year ago
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    furthermore we can note that: \[10 = \sqrt {10} \sqrt {10} \]

  26. Michele_Laino
    • one year ago
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    so we can write: \[\sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \]

  27. Michele_Laino
    • one year ago
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    finally, I factor out sqrt(10) and I get: \[\begin{gathered} \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \hfill \\ = \sqrt {10} \left( {\sqrt {10} \sqrt 5 - 4} \right) \hfill \\ \end{gathered} \]

  28. anonymous
    • one year ago
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    its not too difficult just follow what the question is trying to tell you.

  29. Michele_Laino
    • one year ago
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    recalling taht: \[\sqrt {10} \sqrt 5 = \sqrt {10 \times 5} = \sqrt {50} \] we have: \[\begin{gathered} \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \hfill \\ \hfill \\ = \sqrt {10} \left( {\sqrt {10} \sqrt 5 - 4} \right) = \sqrt {10} \left( {\sqrt {50} - 4} \right) \hfill \\ \end{gathered} \]

  30. Michele_Laino
    • one year ago
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    there is another way to simplify your radical

  31. Michele_Laino
    • one year ago
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    I start from the initial expression: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right)\]

  32. Michele_Laino
    • one year ago
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    and I factor out a 2 inside the parentheses: \[\Large 10 - 4\sqrt 2 = 2\left( {5 - 2\sqrt 2 } \right)\]

  33. Michele_Laino
    • one year ago
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    so I can write: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = \sqrt 5 \cdot 2\left( {5 - 2\sqrt 2 } \right)\]

  34. Michele_Laino
    • one year ago
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    and we have finished

  35. Michele_Laino
    • one year ago
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    now you can choose the method which do you prefer

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