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mathtard

  • 3 years ago

Determine the nature of solutions of the equation x^2-3x+8=0

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  1. fiddlearound
    • 3 years ago
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    you need to check the discriminant - know what that is ?

  2. mathtard
    • 3 years ago
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    no

  3. fiddlearound
    • 3 years ago
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    are you familiar with the quadratic equation ?

  4. mathballet8888
    • 3 years ago
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    use quadratic formula

  5. mathtard
    • 3 years ago
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    no not at all.

  6. fiddlearound
    • 3 years ago
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    ever seen this formula ? \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  7. fiddlearound
    • 3 years ago
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    http://en.wikipedia.org/wiki/Quadratic_equation

  8. mathballet8888
    • 3 years ago
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    \[(-b \pm \sqrt{b^{2}-4ac})/2a\]

  9. fiddlearound
    • 3 years ago
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    In general: ax^2+bx+c=0 In your case: a=1 (the coefficient of x^2) b=-3 (coefficient of x) c=8 Use these values and plug them into the discriminant (the thingy inside the square root). \[\Delta=b^2-4ac\] If the discriminant = 0, there is one real solution If the discriminant > 0, there are two real solutions If the discriminant < 0, there are no real solutions

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