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wendy67

  • 3 years ago

SImplify: Help Please!

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  1. Godfrey_Idrobo_Libreros
    • 3 years ago
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    ???

  2. wendy67
    • 3 years ago
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    \[\frac{ 1+2i }{ \sqrt{2}+i }\]

  3. No-data
    • 3 years ago
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    Multiply both numerator and denominator by the conjugate of the denominator.I mean

  4. No-data
    • 3 years ago
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    \[\sqrt{2}-i\]

  5. No-data
    • 3 years ago
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    \[\frac{1+2i}{\sqrt{2}+i}\cdot \frac{\sqrt{2}-i}{\sqrt{2}-i}\]

  6. wendy67
    • 3 years ago
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    I got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2i ^{2} }{ 2-i ^{2}}\]

  7. No-data
    • 3 years ago
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    Great!

  8. No-data
    • 3 years ago
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    but also you know that \[i^2=-1\]

  9. wendy67
    • 3 years ago
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    yes

  10. wendy67
    • 3 years ago
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    so, it would be: \[\frac{ \sqrt{2}-i+2i \sqrt{2}-2(-1) }{ 2-(-1) }\]

  11. No-data
    • 3 years ago
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    yes, now just make arithmetic operations.

  12. wendy67
    • 3 years ago
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    ok i got: \[\frac{ \sqrt{2}-i+2i \sqrt{2}+2 }{ 3 }\] and this where i got stuck

  13. No-data
    • 3 years ago
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    is there a specific anwer you need to get?

  14. wendy67
    • 3 years ago
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    yes i should get : \[\frac{ 2+\sqrt{2} }{ 3 }+\frac{ 2\sqrt{2}-1 }{ 3 }\]

  15. No-data
    • 3 years ago
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    I got: \[\frac{\sqrt{2}+2}{3}-\frac{1+2\sqrt{2}}{3}i\]

  16. wendy67
    • 3 years ago
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    ok so how did you get that?

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