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madelyn97

  • 2 years ago

If 3+2i is a solution for x^2+mx+n , where m and n are real numbers, what is the value of m?

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  1. satellite73
    • 2 years ago
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    easier than you think

  2. satellite73
    • 2 years ago
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    if \(3+2i\) is a zero so is \(3-2i\) the quadratic factors as \[(x-(3+2i))(x-(3-2i))\] when you multiply out you will get \(x^2\) as the first term last term \(n\) will be \(3+2i)(3-2i)=3^2+2^2=14\)and the "middle term" will be\[-(3+2i)x-(3-2i)x=-6x\]

  3. satellite73
    • 2 years ago
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    actually \(3^2+2^2=13\) but whatever

  4. satellite73
    • 2 years ago
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    the original quadratic was \[x^2-6x+13\]

  5. alanli123
    • 2 years ago
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    m should be -6

  6. satellite73
    • 2 years ago
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    once you see this you can know if \(a+bi\) is a zero, s is \(a-bi\) and the original quadratic is \[x^2-2ax+a^2+b^2\]

  7. satellite73
    • 2 years ago
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    btw you can always work backwards: \[x=3+2i\] \[x-3=-2i\] \[(x-3)^2=(-2i)^2=-4\] \[(x-3)^2+4=0\] \[x^2-6x+9+4=0\]

  8. madelyn97
    • 2 years ago
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    okay so how do I get the value of m from there?

  9. satellite73
    • 2 years ago
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    \[x^2+mx+n=0\] \[x^2-6x+13=0\] what is \(m\)?

  10. madelyn97
    • 2 years ago
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    the original question asked for the value of m

  11. madelyn97
    • 2 years ago
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    so m is -6?

  12. madelyn97
    • 2 years ago
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    Thanks!!

  13. cinar
    • 2 years ago
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    \[\Large(3+2i)^2+m(3+2i)+n=0\] \[\Large 9+12i+4i^2+3m+2 mi+n=0\] \[\Large (3m+n)+ 2mi=-12i-5\] \[\Large 3m+n=-5 \\2m=-12\\m=-6\\n=13\]

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