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anonymous

  • one year ago

What indicators predict that a quadratic funtion will have a complex solution? [Solve and show all work] x^2 - 10x = -13 x = -b +- squareroot b^2 -4ac OVER 2a

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  1. IrishBoy123
    • one year ago
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    focus on "b^2 -4ac " as you need to take its square root.....

  2. anonymous
    • one year ago
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    Well, I'm not sure how to solve it. Can you teach me? @IrishBoy123

  3. IrishBoy123
    • one year ago
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    what is b and what is c in your original quadratic?

  4. IrishBoy123
    • one year ago
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    this is the quadratic you posted: \(\large x^2 - 10x = -13 \) and this is the additional formula you also posted: \(\large x = \frac{-b \ \pm \ \sqrt{ b^2 -4ac} }{ 2a}\)

  5. IrishBoy123
    • one year ago
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    maybe it would be better to start by asking you what you are trying to do to the quadratic? i know but do you also know?

  6. anonymous
    • one year ago
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    No, I'm unsure on how to solve this question...

  7. IrishBoy123
    • one year ago
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    that's my question....you are trying to find the value of x where the quadratic = 0, is that right?

  8. anonymous
    • one year ago
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    Well, the question is asking to find complex solutions... Which I dont really understand what it means by that.

  9. anonymous
    • one year ago
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    and I guess it's asking to solve also.. So yeah. solving for x

  10. IrishBoy123
    • one year ago
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    take your original equation \(x^2 - 10x = -13 \) and re-write it as \(x^2 - 10x +13 =0\) , yep?!! then compare it to this \(a x^2 + b x + c = 0\) and then apply that second equation you posted, the one with "\(b^2 - 4 ac\)" in it [i set it out again above] that second equation gives you the roots (x intercepts) of the quadratic, ie x values for when quadratic = 0 try it and see where you go.....

  11. IrishBoy123
    • one year ago
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    clearly if \(b^2 - 4ac \lt 0\) you might have an issue (ie complex roots) but run with it first....

  12. anonymous
    • one year ago
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    Well, I'm sorta confused on how to plug it in and solve it... Can you help me on that?

  13. anonymous
    • one year ago
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    @IrishBoy123 you there?

  14. IrishBoy123
    • one year ago
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    compare \(x^2−10x+13=0\) to \(ax^2+bx+c=0\) so \(a = 1, b = -10, c = 13\) plug these into \(\huge x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

  15. IrishBoy123
    • one year ago
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    what do you get?

  16. anonymous
    • one year ago
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    |dw:1438365369887:dw|

  17. anonymous
    • one year ago
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    okay, I regret not paying enough attention in math class... How do I solve this? Sorry to take up more of your time by asking this. I never got anything out of quadratics... *sigh*

  18. IrishBoy123
    • one year ago
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    cool, but its \(-b \pm\) and -(-10) = 10 but now get out your calculator/spreadsheet/whatever and calculate the 2 values of x. it is \(\pm\) so there will be 2 values. these are your roots

  19. anonymous
    • one year ago
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    Okay.

  20. IrishBoy123
    • one year ago
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    \(\huge \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(13)}}{2(1)} \implies \) \(\huge \frac{10 \pm \sqrt{100 - 52}}{2} = 1.54, \ 8.46 \)

  21. IrishBoy123
    • one year ago
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    yep?

  22. anonymous
    • one year ago
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    Yeah. Thank you.

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