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BazBendell

  • 2 years ago

Simplify: (Points : 1) 3

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  1. BazBendell
    • 2 years ago
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  2. Hero
    • 2 years ago
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    \[\text{Let} \space \sqrt{2} = x \space \text{then} \space 6\sqrt{2} - 3\sqrt{2} = 6x - 3x = 3x = 3\sqrt{2}\]

  3. imagreencat
    • 2 years ago
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    Basically, you are just going to combine like terms. And when one term has a sqrt(2), and there's a value before it, we can take 3sqrt(2) + 2sqrt(2) to be equal to 5sqrt(2). It's like saying 3 cows + 2 cows = 5 cows. the sqrt(2) is like the label of each term. Actually, like what Hero said above. Now, you may be confused with the second and third numbers because they contain two labels now. Don't mix them, because sqrt(y) is not equal to the sqrt(x) of the sqrt(z). These are three different labels. Just leave them that way. Like the expressing 3x + 2y = 5z is left that way.

  4. Chlorophyll
    • 2 years ago
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    Do you still need help?

  5. BazBendell
    • 2 years ago
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    I have no idea what either of you said...

  6. Chlorophyll
    • 2 years ago
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    When you add or subtract the terms with root, the best way is factorize the root out!

  7. Chlorophyll
    • 2 years ago
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    Do you understand my explanation?

  8. Chlorophyll
    • 2 years ago
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    For example: 6√2 - 3√2 = ( 6 -3 ) *√2 = 3 √2

  9. Chlorophyll
    • 2 years ago
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    Of course, only the terms have the same root can be factored!

  10. BazBendell
    • 2 years ago
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    Not a clu what that means.

  11. Chlorophyll
    • 2 years ago
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    6√2 - 3√2 Which part do you see the same?

  12. Hero
    • 2 years ago
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    I hope @BazBendell isn't a first grader trying to understand 7th grade work.

  13. Chlorophyll
    • 2 years ago
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    @BazBendell would you prove that Hero is completely wrong ;)

  14. Hero
    • 2 years ago
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    The only way he could prove me wrong is by understanding the material somehow ;)

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spraguer (Moderator)
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