# How do you find the roots, real and imaginary, of y=4x^2+12x-25 using the quadratic formula?

Mar 7, 2018

#### Answer:

$x = \frac{- 3 \pm \sqrt{34}}{2}$

#### Explanation:

$4 {x}^{2} + 12 x - 25 = a {x}^{2} + b x + c$

set $4 {x}^{2} + 12 x - 25$ to 0 to find the roots real and imaginary

Using the discriminant ${b}^{2} - 4 a c$ to find out if there are imaginary roots

${\left(12\right)}^{2} - 4 \left(4\right) \left(- 25\right) = 544$

$\Delta \left(\text{the discriminant}\right) > 0$ So that means it have 2 unique real roots

Using the quadratic formula

$x = \frac{- \left(12\right) \pm \sqrt{{\left(12\right)}^{2} - 4 \left(4\right) \left(- 25\right)}}{2 \left(4\right)}$

$x = \frac{- 12 \pm \sqrt{544}}{8}$

simplify the radical using $\sqrt[n]{x} y = \sqrt[n]{x} \cdot \sqrt[n]{y}$

$x = \frac{- 12 \pm \sqrt{16 \cdot 34}}{8}$

$x = \frac{- 12 \pm \sqrt{4} \sqrt{34}}{8}$

$x = \frac{- 12 \pm 2 \sqrt{34}}{8}$

Find the greatest common factor

$x = \frac{- 3 \pm \sqrt{34}}{2}$

$x = \frac{- 3 + \sqrt{34}}{2}$ and $- \frac{3 + \sqrt{34}}{2}$

Using $C . T . S .$ (complete the square)
To get ${x}^{2}$ by itself divide the whole equation by $4$

$\frac{4 {x}^{2} + 12 x - 25}{4} = \frac{0}{4}$

$\frac{\cancel{4} {x}^{2}}{\cancel{4}} + \frac{3 \cancel{12} x}{1 \cancel{4}} - \frac{25}{4} = 0$

${x}^{2} + 3 x \cancel{- \frac{25}{4} + \frac{25}{4}} = 0 + \frac{25}{4}$

${x}^{2} + 3 x = \frac{25}{4}$

complete the square by dividing the coefficient of x by 2 and then squaring it

${\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$

add this term to both sides

${x}^{2} + 3 x + \frac{9}{4} = \frac{25}{4} + \frac{9}{4}$

sqrt((x+3/2)^2)=+-sqrt(34/4

$x \cancel{+ \frac{3}{2} - \frac{3}{2}} = \frac{\pm \sqrt{34}}{2} + \frac{- 3}{2}$

$x = \frac{- 3 \pm \sqrt{34}}{2}$

$\therefore$ $x = \frac{- 3 + \sqrt{34}}{2}$ and $\frac{- 3 - \sqrt{34}}{2}$