# How do you use the quadratic formula to find the zeros of the function 12x^2 - 5x - 7 = 0?

Jul 29, 2017

Zeros of the function are $x = 1 \mathmr{and} x = - \frac{7}{12}$

#### Explanation:

$12 {x}^{2} - 5 x - 7 = 0 \mathmr{and} 12 {x}^{2} - 12 x + 7 x - 7 = 0$ or

$12 x \left(x - 1\right) + 7 \left(x - 1\right) = 0 \mathmr{and} \left(x - 1\right) \left(12 x + 7\right) = 0$

Zeros are either $x - 1 = 0 \therefore x = 1$ OR $12 x + 7 = 0 \therefore x = - \frac{7}{12}$

Zeros of the function are $x = 1 \mathmr{and} x = - \frac{7}{12}$ [Ans]

Jul 29, 2017

zeroes: $x \in \left\{- \frac{7}{12} , 1\right\}$

#### Explanation:

The quadratic formula tells us us that if an equation has the form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
then it has zeroes:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{12} {x}^{2} \textcolor{b l u e}{- 5} x \textcolor{g r e e n}{- 7} = 0$
has zeroes:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{\left(- 5\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 5\right)}}^{2} - 4 \cdot \textcolor{red}{12} \cdot \textcolor{g r e e n}{\left(- 7\right)}}}{2 \cdot \textcolor{red}{12}}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{5 \pm \sqrt{25 + 336}}{24}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{5 \pm 19}{24}$

color(white)("XXXX")= 1 " or " -7/12

Jul 29, 2017

Zeros are $x = 1 \mathmr{and} x = - \frac{7}{12}$

#### Explanation:

 x = - -b/(2a) +- sqrt(b^2-4ac)/(2a ); a=12 , b= -5 ,c =-7 ;

$D = {b}^{2} - 4 a c = 25 + 336 = 361 \therefore x = \frac{5}{24} \pm \frac{19}{24}$

$x = \frac{5}{24} + \frac{19}{24} \mathmr{and} x = 1 \mathmr{and} x = \frac{5}{24} - \frac{19}{24} = - \frac{14}{24} = - \frac{7}{12}$

Zeros are $x = 1 \mathmr{and} x = - \frac{7}{12}$ [Ans]