# How do you find the zeros, real and imaginary, of y=-x^2+4x+12 using the quadratic formula?

Apr 21, 2018

See a solution process below:

#### Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{- 1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{4}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{12}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{{\textcolor{b l u e}{4}}^{2} - \left(4 \cdot \textcolor{red}{- 1} \cdot \textcolor{g r e e n}{12}\right)}}{2 \cdot \textcolor{red}{- 1}}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{16 - \left(- 48\right)}}{- 2}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{16 + 48}}{- 2}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{64}}{- 2}$

$x = \frac{- \textcolor{b l u e}{4} - 8}{- 2}$ and $x = \frac{- \textcolor{b l u e}{4} + 8}{- 2}$

$x = \frac{- 12}{- 2}$ and $x = \frac{4}{- 2}$

$x = 6$ and $x = - 2$

Apr 21, 2018

$f \left(x\right) = a {x}^{2} + b x + c$
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

#### Explanation:

$f \left(x\right) = - {x}^{2} + 4 x + 12 = 0$
$a = - 1 , b = 4 , c = 12$

${x}_{1} = \setminus \frac{- 4 + \setminus \sqrt{{4}^{2} - 4 \cdot \left(- 1\right) \cdot 12}}{- 2} =$

$= \setminus \frac{- 4 + \setminus \sqrt{16 + 48}}{- 2} = \frac{- 4 + \setminus \sqrt{64}}{- 2} = \frac{- 4 + 8}{- 2} = - 2$

${x}_{2} = \setminus \frac{- 4 - \setminus \sqrt{{4}^{2} - 4 \cdot \left(- 1\right) \cdot 12}}{- 2} =$

$= \setminus \frac{- 4 - \setminus \sqrt{16 + 48}}{- 2} = \frac{- 4 - \setminus \sqrt{64}}{- 2} = \frac{- 4 - 8}{- 2} = 6$

In this case all zeros are real.

A polynomial of degree n has n zeros. Imaginary zeros are always in conjugate pairs.
$f \left(x\right) = {x}^{2} + 10 x + 169$
${x}^{2} + 10 x + 169 = 0$
$a = 1 , b = 10 , c = 169$
${x}_{1 , 2} = \setminus \frac{- 10 \setminus \pm \setminus \sqrt{{10}^{2} - 4 \cdot 1 \cdot 169}}{2} =$
$= \setminus \frac{- 10 \setminus \pm \setminus \sqrt{100 - 676}}{2} =$

$= \setminus \frac{- 10 \setminus \pm \setminus \sqrt{- 576}}{2}$

Remember: $i \cdot i = - 1 , \setminus \sqrt{- 1} = i$
$\setminus \sqrt{- 576} = \setminus \sqrt{- {2}^{6} \cdot {3}^{2}} = {2}^{3} \cdot 3 \cdot i$

${x}_{1 , 2} = \setminus \frac{- 10 \setminus \pm \setminus \sqrt{- 576}}{2} =$
$= \setminus \frac{- 10 \setminus \pm {2}^{3} \cdot 3 \cdot i}{2} =$
$= - 5 \setminus \pm {2}^{2} \cdot 3 \cdot i = - 5 \setminus \pm 12 \cdot i$
$\implies$ since complex zeros are always in pairs, a polynomial of degree $2$ has or real or complex zeros.

( a polynomial of degree $3$ has:
or $3$ real zeros
or $1$ real and $2$ complex zeros.)