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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 ?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can always use the quadratic formula for solving a quadratic equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0expression=0 right side has to =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0completing 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2+6x1=0\] 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok not done. \[x=3+\sqrt{10}\] \[x=3\sqrt{10}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now by quadratic formula: a = 1, b = 6, c = 1 \[x= \frac{6 + \sqrt{6^24\times 1 \times 1}}{2}\] \[x=\frac{6 + \sqrt{36+4}}{2}\] \[x=\frac{6 + \sqrt{40}}{2}\] \[x=\frac{6 + \sqrt{4\times10}}{2}\] \[x=\frac{6 +2 \sqrt{10}}{2}\] \[x=\frac{2(3 + \sqrt{10})}{2}\] \[x=3+\sqrt{10}\] what a pain
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