# Given -2x^2+16x-26 how do you use the quadratic formula to find the zeros of f(x)?

Nov 1, 2016

$x = 4 + \sqrt{3}$
Or
$x = 4 - \sqrt{3}$

#### Explanation:

FIRST METHOD:

The zeros of the given polynomial is determined by completing the square and applying polynomial properties

$\textcolor{b r o w n}{{\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}}$
$\textcolor{v i o \le t}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

$- 2 {x}^{2} + 16 x - 26$

$= \textcolor{b l u e}{- 2} \left(\textcolor{b l u e}{1} {x}^{2} \textcolor{b l u e}{- 8} x \textcolor{b l u e}{+ 13}\right)$

$= - 2 \left({x}^{2} - 8 x + 13 \textcolor{red}{+ 3 - 3}\right)$

$= - 2 \left({x}^{2} - 8 x + 16 - 3\right)$

$= - 2 \left(\textcolor{b r o w n}{{x}^{2} - 2 \left(4\right) x + {4}^{2}} - 3\right)$

=-2(color(brown)((x-4)^2)-3)^

$= - 2 \textcolor{v i o \le t}{\left({\left(x - 4\right)}^{2} - {\left(\sqrt{3}\right)}^{2}\right)}$

$= - 2 \textcolor{v i o \le t}{\left(\left(x - 4\right) - \sqrt{3}\right) \left(\left(x - 4\right) + \sqrt{3}\right)}$

=-2((x-(4+sqrt3)((x-(4-sqrt3))

The zeros of the given algebraic expression is

$- 2 {x}^{2} + 16 x - 26 = 0$

-2((x-(4+sqrt3)((x-(4-sqrt3))=0

$x - \left(4 + \sqrt{3}\right) = 0 \Rightarrow x = 4 + \sqrt{3}$
Or
$x - \left(4 - \sqrt{3}\right) = 0 \Rightarrow x = 4 - \sqrt{3}$

SECOND METHOD:

The zeros of the given polynomial is determined by using the quadratic formula

$- 2 {x}^{2} + 16 x - 26$

$\textcolor{b l u e}{\delta = {b}^{2} - 4 a c} = {16}^{2} - 4 \left(- 2\right) \left(- 26\right)$
$\delta = 256 - 208 = 48$

Roots are:
$\textcolor{red}{{x}_{1} = \frac{- b - \sqrt{\delta}}{2 a}} = \frac{- 16 - \sqrt{48}}{2 \left(- 2\right)} = \frac{- 16 - \sqrt{{2}^{4} \times 3}}{- 4}$
$\textcolor{red}{{x}_{1}} = \frac{- 16 - 4 \sqrt{3}}{- 4} = \frac{- 4 \left(4 + \sqrt{3}\right)}{- 4} = \frac{\cancel{- 4} \left(4 + \sqrt{3}\right)}{\cancel{- 4}}$

$\textcolor{red}{{x}_{1}} = 4 + \sqrt{3}$

$\textcolor{red}{{x}_{2} = \frac{- b + \sqrt{\delta}}{2 a}} = \frac{- 16 + \sqrt{48}}{2 \left(- 2\right)} = \frac{- 16 + \sqrt{{2}^{4} \times 3}}{- 4}$
$\textcolor{red}{{x}_{2}} = \frac{- 16 + 4 \sqrt{3}}{- 4} = \frac{- 4 \left(4 - \sqrt{3}\right)}{- 4} = \frac{\cancel{- 4} \left(4 - \sqrt{3}\right)}{\cancel{- 4}}$

$\textcolor{red}{{x}_{2}} = 4 - \sqrt{3}$