How do you solve by completing the square: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ?

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How do you solve by completing the square: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ?

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Answer 1 · 2015-04-01T01:40:05

$a x^{2} + b x + c = 0$ $ax^2 + bx +c = 0$

Divide all terms by $a$ $a$ so as to reduce the coefficient of $x^{2}$ $x^2$ to $1$ $1$
$x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$ $x^2 +b/a x + c/a=0$

Subtract the constant term from both sides of the equation
$x^{2} + \frac{b}{a} x = - \frac{c}{a}$ $x^2+b/a x = - c/a$

To have a square on the left side the third term (constant) should be
${(\frac{b}{2 a})}^{2}$ $(b/(2a))^2$

So add that amount to both sides
$x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} = {(\frac{b}{2 a})}^{2} - \frac{c}{a}$ $x^2 +b/a x +(b/(2a))^2 = (b/(2a))^2 - c/a$

Re-write the left-side as a square
${(x + (\frac{b}{2 a}))}^{2} = {(\frac{b}{2 a})}^{2} - \frac{c}{a}$ $(x+(b/(2a)))^2 = (b/(2a))^2 -c/a$

Take the square root of both sides (remembering that the result could be plus or minus)
$x + (\frac{b}{2 a}) = \pm \sqrt{{(\frac{b}{2 a})}^{2} - \frac{c}{a}}$ $x+(b/(2a)) = +-sqrt((b/(2a))^2 -c/a)$

Subtract the constant term on the left side from both sides
$x = \pm \sqrt{{(\frac{b}{2 a})}^{2} - \frac{c}{a}} - (\frac{b}{2 a})$ $x = +-sqrt((b/(2a))^2 -c/a) - (b/(2a))$

or, with some simplification

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$

(the standard form for solving a quadratic)

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How do you solve by completing the square: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ?

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How do you solve by completing the square: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ ?