# How do you solve x^2 + 4x - 21 = 0?

Jun 1, 2015

We are going to solve it with $\Delta$

The equation is of the form ax²+bx+c=0

$a = 1$ $b = 4$ $c = - 21$

Delta=b²-4ac
Delta=4²-4*1*(-21)
$\Delta = 16 + 84$
$\Delta = 100$ ( $= \sqrt{10}$)

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{- 4 - 10}{2}$

${x}_{1} = - \frac{14}{2}$

${x}_{1} = - 7$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$

${x}_{2} = \frac{- 4 + 10}{2}$

${x}_{2} = \frac{6}{2}$

${x}_{2} = 3$

Jun 1, 2015

You can solve this by using the general formula for complete quadratic equations.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
For a complete quadratic equation $a {x}^{2} + b x + c = 0$.
Let's apply it to your equation, which gives:

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot \left(- 21\right)}}{2} = \frac{- 4 \pm \sqrt{100}}{2} = \frac{- 4 \pm 10}{2} = - 2 \pm 5$
So, we've got two solutions:
$x = 3$
and
$x = - 7$

Jun 1, 2015

A less direct, but perhaps simpler approach is to recognise that if ${x}^{2} + 4 x - 21$ has linear factors, then they must look like:

$\left(x + p\right) \left(x - q\right) = {x}^{2} + \left(p - q\right) x - \left(p \times q\right)$ where $p , q > 0$

The reason I say it must look like this with one $+$ and one $-$ is because the sign of the constant term in the original quadratic is negative.

Then $p - q = 4$ and $p \times q = 21$.

What pair of factors of $21$ differs by $4$?

There are not many choices of factorizations, basically either $1 \times 21$ or $3 \times 7$.

$7 - 3 = 4$ and $7 \times 3 = 21$, so let $p = 7$ and $q = 3$

${x}^{2} + 4 x - 21 = \left(x + p\right) \left(x - q\right) = \left(x + 7\right) \left(x - 3\right)$

This is zero when $x = - 7$ or $x = 3$