How do I find all real and complex zeros of x^2-4x+7?

Oct 31, 2015

Use the quadratic formula to find zeros occur for:

$x = 2 \pm i \sqrt{3}$

Explanation:

The quickest way is probably to use the quadratic formula.

${x}^{2} - 4 x + 7$ is in the form $a {x}^{2} + b x + c$, with $a = 1$, $b = - 4$ and $c = 7$.

So its zeros occur for:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \times 1 \times 7\right)}}{2 \times 1}$

$= \frac{4 \pm \sqrt{16 - 28}}{2} = \frac{4 \pm \sqrt{- 12}}{2} = \frac{4 \pm i \sqrt{12}}{2}$

$= \frac{4 \pm i \cdot 2 \sqrt{3}}{2} = 2 \pm i \sqrt{3}$

Alternatively, you can complete the square as follows:

$0 = {x}^{2} - 4 x + 7 = {x}^{2} - 4 x + 4 + 3 = {\left(x - 2\right)}^{2} + 3$

Subtract $3$ from both ends to get:

${\left(x - 2\right)}^{2} = - 3$

So:

$x - 2 = \pm \sqrt{- 3} = \pm i \sqrt{3}$

Add $2$ to both sides to get:

$x = 2 \pm i \sqrt{3}$