How do you find all the real and complex roots of x^2-2x-5=0?

Jun 10, 2016

You find them applying the formula for the second order equations.

Explanation:

Every second order equation in the form

$a {x}^{2} + b x + c = 0$ can be solved applying the formula

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

The roots are two, one when you use the sign $+$ and one with the sign $-$.
So the only thing that you have to do is to identify $a , b , c$ in your equation and substitute in the formula.

${x}^{2} - 2 x - 5 = 0$ and if you confront it with
$a {x}^{2} + b x + c = 0$ you see that

$a = 1$, $b = - 2$ and $c = - 5$.

Then the solution is

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

$= \frac{2 \setminus \pm \sqrt{\left(4 + 8\right)}}{2}$

$= \frac{2 \setminus \pm \sqrt{12}}{2}$, I use the fact that $\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$

$= \frac{2 \setminus \pm 2 \sqrt{3}}{2}$

$= 1 \setminus \pm \sqrt{3}$

The two roots are

${x}_{1} = 1 + \sqrt{3} \setminus \approx 2.73$
${x}_{2} = 1 - \sqrt{3} \setminus \approx - 0.73$.