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angelwings996

  • 2 years ago

What is the number of real solutions? -11x^2 = x + 11

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  1. dumbcow
    • 2 years ago
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    first set equal to zero second find discriminate (b^2 -4ac) third, if discriminate >0 : 2 real solutions if discriminate =0: 1 real solution if discriminate <0 : no real solutions

  2. angelwings996
    • 2 years ago
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    would the equation be -11x^2 - x - 11 = 0 ?

  3. dumbcow
    • 2 years ago
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    yes or you can get rid of the neg --> 11x^2 +x+11 = 0

  4. angelwings996
    • 2 years ago
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    Okay, now what would I do?

  5. dumbcow
    • 2 years ago
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    find Discriminant http://en.wikipedia.org/wiki/Discriminant

  6. hartnn
    • 2 years ago
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    Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then calculate \(b^2-4ac=...?\)

  7. angelwings996
    • 2 years ago
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    don't I already have a, b, and c so would I just substitute them into b^2 - 4ac ?

  8. hartnn
    • 2 years ago
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    what u got, a=..? b=...?c=..?

  9. angelwings996
    • 2 years ago
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    doesn't a = 11, b = 1, and c = 11 ?

  10. hartnn
    • 2 years ago
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    thats correct, now as @dumbcow said, u just need to find whether b^2-4ac is positive or negative...

  11. angelwings996
    • 2 years ago
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    so if I plug it in would it be like this .... \[1 ^{2} - 4(11)(11)\]

  12. hartnn
    • 2 years ago
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    yes, go on.

  13. angelwings996
    • 2 years ago
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    okay so \[1^{2} = 1 and -4(11) = -44(11) = -484\] Then you would subtract -484 from 1 which gives you -483, right ?

  14. hartnn
    • 2 years ago
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    yes, so how many real solutions ?

  15. angelwings996
    • 2 years ago
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    since -483 is less than 0 would it be no real solutions?

  16. hartnn
    • 2 years ago
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    absolutely correct :)

  17. angelwings996
    • 2 years ago
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    Okay thank you so much for both of your helps!

  18. hartnn
    • 2 years ago
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    welcome ^_^

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