# How do you solve x^2+4x-3=0 using the quadratic formula?

Apr 19, 2016

Use the quadratic formula to find:

$x = - 2 \pm \sqrt{7}$

#### Explanation:

${x}^{2} + 4 x - 3 = 0$ is of the form $a {x}^{2} + b x + c = 0$, with $a = 1$, $b = 4$ and $c = - 3$.

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 4 \pm \sqrt{{4}^{2} - \left(4 \cdot 1 \cdot \left(- 3\right)\right)}}{2 \cdot 1}$

$= \frac{- 4 \pm \sqrt{16 + 12}}{2}$

$= \frac{- 4 \pm \sqrt{28}}{2}$

$= \frac{- 4 \pm \sqrt{{2}^{2} \cdot 7}}{2}$

$= \frac{- 4 \pm 2 \sqrt{7}}{2}$

$= - 2 \pm \sqrt{7}$

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Using the quadratic formula is similar in some ways to completing the square:

$0 = {x}^{2} + 4 x - 3$

$= {x}^{2} + 4 x + 4 - 7$

$= {\left(x + 2\right)}^{2} - {\left(\sqrt{7}\right)}^{2}$

$= \left(\left(x + 2\right) - \sqrt{7}\right) \left(\left(x + 2\right) + \sqrt{7}\right)$

$= \left(x + 2 - \sqrt{7}\right) \left(x + 2 + \sqrt{7}\right)$

Hence $x = - 2 \pm \sqrt{7}$

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In fact, we can derive the quadratic formula as follows:

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

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

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

Hence:

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

Divide both sides by $a$ to get:

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

Hence:

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

So:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$