# How do you find the zeros, real and imaginary, of y=-x^2-2x+11 using the quadratic formula?

Nov 20, 2015

This equation has 2 real (irrational) solutions.

#### Explanation:

To find the zeros of a quadratic equation you use a formula:

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

$\Delta = {b}^{2} - 4 a c$

This zeroes are:

• real and different $\iff \Delta > 0$
• real and equal $\iff \Delta = 0$
• conjugate complex numbers $\iff \Delta < 0$

To calculate ${x}_{1 , 2}$ in the last case you have to remember, that if $a < 0$ then $\sqrt{a} = \sqrt{\left\mid a \right\mid} \cdot i$, where $i$ is an imaginary unit.

If we use this rule for our function we get:

$\Delta = 4 - 4 \cdot \left(- 1\right) \cdot 11$

$\Delta = 4 + 44 = 48$

In this place we can say that the roots of the equation are real (and irrational) because $48 > 0$ (and $\sqrt{48}$ is not a natural number).

Now we are looking for ${x}_{1}$ and ${x}_{2}$

${x}_{1} = \frac{- b + \sqrt{\Delta}}{2 a} = \frac{2 - 4 \sqrt{3}}{- 2} = - 1 + 2 \sqrt{3}$

${x}_{2} = \frac{- b - \sqrt{\Delta}}{2 a} = \frac{2 + 4 \sqrt{3}}{- 2} = - 1 - 2 \sqrt{3}$

Note that we can say that this function has complex zeros (because all real numbers are also complex $\mathbb{R} \subset \mathbb{C}$), but we cannot say that the zeros are imaginary (because an imaginary number is a complex number with real part equal to zero according to http://mathworld.wolfram.com/ImaginaryNumber.html)