What is the quadratic formula?

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What is the quadratic formula?

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Answer 1 · 2018-03-26T02:41:37

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

Explanation:

Negative b plus minus the square root of b squared minus 4*a*c over 2*a. To plug something into the quadratic formula the equation needs to be in standard form ( $a x^{2} + b x^{2} + c$ $ax^2 + bx^2 +c$ ).

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Answer 2 · 2018-03-27T00:25:44

If we have:

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

Then:

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

Explanation:

The quadratic formula provides a method of solving a generic quadratic equation:

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

To solve the equation we first factor out $a$ $a$ :

$a {x^{2} + \frac{b}{a} x + \frac{c}{a}} = 0 \Rightarrow x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$ $a{x^2 + b/ax + c/a} = 0 => x^2 + b/ax + c/a = 0$

Then we complete the square:

${(x + \frac{b}{2 a})}^{2} - {(\frac{b}{2 a})}^{2} + \frac{c}{a} = 0$ $(x + b/(2a))^2 - (b/(2a))^2 + c/a = 0$

Now, we solve for $x$ $x$ :

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

$= \frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}}$ $" " = b^2/(4a^2) - (4ac)/(4a^2)$

$= \frac{b^{2} - 4 a c}{4 a^{2}}$ $" " = (b^2-4ac)/(4a^2)$

By taking square root we get:

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

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

So that:

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

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

Which is known as the "quadratic formula".

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What is the quadratic formula?

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