# How do you find the value of the discriminant and determine the nature of the roots -4r^2-4r=6?

Feb 8, 2017

We have complex conjugate roots for the given equation.

#### Explanation:

The discriminant of quadratic equation $a {x}^{2} + b x + c = 0$ is $\Delta = {b}^{2} - 4 a c$ and roots are given by $\frac{- b \pm \sqrt{\Delta}}{2 a}$

If $\Delta$ is positive and a square of a rational number (and so are $a$ and $b$), roots are rational .

If $\Delta$ is positive but not a square of a rational number, roots are irrational and real .

If $\Delta = 0$, we have only one root given by $\frac{- b}{2 a}$.

If $\Delta$ is negative but $a$, $b$ and $c$ are real numbers, roots are two complex conjugate. But if $\Delta$ is negative but $a$, $b$ are not real numbers, roots are complex .

Hence, one needs to first convert the equation $- 4 {r}^{2} - 4 r = 6$ in this form. This can be easily done by shifting terms on left hand side to RHS and this become $0 = 4 {r}^{2} + 4 r + 6$ or $4 {r}^{2} + 4 r + 6 = 0$

As we have $a = 4$, $b = 4$ and $c = 6$,

the determinant is ${4}^{2} - 4 \times 4 \times 6 = 16 - 96 = - 80$

As the determinant is negative and $a$, $b$ and $c$ are real numbers, we have complex conjugate roots for the given equation.

These are $\frac{- 4 \pm \sqrt{- 80}}{2 \times 4} = - \frac{1}{2} \pm i \frac{\sqrt{5}}{2}$

i.e. $- \frac{1}{2} - i \frac{\sqrt{5}}{2}$ and $- \frac{1}{2} + i \frac{\sqrt{5}}{2}$