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daze

  • 3 years ago

Solve x2 + 4x + 8 = 0

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  1. BluFoot
    • 3 years ago
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    You have to use the quadratic equation, do you know it?

  2. daze
    • 3 years ago
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    no

  3. BluFoot
    • 3 years ago
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    \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  4. nickersia
    • 3 years ago
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    ax2+bx+c=0 Formula: (-b+-sqrt(b squared -4*a*c))/2a

  5. daze
    • 3 years ago
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    so?

  6. BluFoot
    • 3 years ago
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    a=1, b=4, c=8, plug 'em in

  7. daze
    • 3 years ago
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    x equals negative 2 plus or minus 4 I?

  8. daze
    • 3 years ago
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    BluFoot?

  9. nickersia
    • 3 years ago
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    +- stands for two possible resoults, both are correct

  10. jazy
    • 3 years ago
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    \[ax^2 + bx + c = 0\]\[x^2 + 4x + 8 = 0\]a = 1 b = 4 c = 8 \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a } \] As BluFoot mentioned, just plug them in.\[x=\frac{ -(4) \pm \sqrt{(4)^2-4(1)(8)} }{ 2(1) } \]

  11. jazy
    • 3 years ago
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    Now lets simplify that a bit. \[x=\frac{ -(4) \pm \sqrt{(4)^2-4(1)(8)} }{ 2(1)} \] \[x=\frac{ -(4) \pm \sqrt{16-32} }{ 2 }\]

  12. daze
    • 3 years ago
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    x equals negative 4 plus or minus 4 I?

  13. daze
    • 3 years ago
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    jazy?

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