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anonymous

  • 5 years ago

when do we use quadratic formula ? Can we use it whenever we want or there are special conditions when we use them ?

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  1. anonymous
    • 5 years ago
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    you can always use the quadratic formula for solving a quadratic equation.

  2. anonymous
    • 5 years ago
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    expression=0 right side has to =0

  3. anonymous
    • 5 years ago
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    completing the square is usually easiest when the leading coefficient is 1 and the 'middle term' has even coefficient, like the example \[x^2+4x=21\] the quadratic equation forces a denominator on you which is often unnecessary. factoring is easiest but only works if the problem is cooked up to factor, like in an elementary algebra text.

  4. anonymous
    • 5 years ago
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    \[x^2+6x-1=0\] via completing the square: \[x^2+6x=1\] \[(x+3)^2=1+3^2=10\] \[x+3=\sqrt{10}\] or \[x+3=-\sqrt{10}\] done

  5. anonymous
    • 5 years ago
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    ok not done. \[x=-3+\sqrt{10}\] \[x=-3-\sqrt{10}\]

  6. anonymous
    • 5 years ago
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    now by quadratic formula: a = 1, b = 6, c = -1 \[x= \frac{-6 + \sqrt{6^2-4\times 1 \times -1}}{2}\] \[x=\frac{-6 + \sqrt{36+4}}{2}\] \[x=\frac{-6 + \sqrt{40}}{2}\] \[x=\frac{-6 + \sqrt{4\times10}}{2}\] \[x=\frac{-6 +2 \sqrt{10}}{2}\] \[x=\frac{2(-3 + \sqrt{10})}{2}\] \[x=-3+\sqrt{10}\] what a pain

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