# How do you find the roots, real and imaginary, of y= x^2 + 4x - 3  using the quadratic formula?

Feb 29, 2016

See explanation...

#### Explanation:

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

This has zeros given by the quadratic formula:

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

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

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

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

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

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

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

Alternative Method

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Complete the square and use this with $a = x + 2$ and $b = \sqrt{7}$ as follows:

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

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

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

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

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

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