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

1 Answer
Feb 8, 2017

Answer:

We have complex conjugate roots for the given equation.

Explanation:

The discriminant of quadratic equation #ax^2+bx+c=0# is #Delta=b^2-4ac# and roots are given by #(-b+-sqrtDelta)/(2a)#

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 #(-b)/(2a)#.

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 #-4r^2-4r=6# in this form. This can be easily done by shifting terms on left hand side to RHS and this become #0=4r^2+4r+6# or #4r^2+4r+6=0#

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

the determinant is #4^2-4xx4xx6=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 #(-4+-sqrt(-80))/(2xx4)=-1/2+-isqrt5/2#

i.e. #-1/2-isqrt5/2# and #-1/2+isqrt5/2#