Find the discriminant of this quadratic equation then state the number and type of solutions. 6b^2+b+3= -5

Expert Answers
rakesh05 eNotes educator| Certified Educator

Because given equation is a quadratic equation i.e. of degree 2. So, this equation will always have two solutions (either real or complex).

The discriminant for the quadratic  equation  `px^2+qx+r=0`

is given by    `D=q^2-4pr` .

Now if (i) `D>0` , both roots are real and distinct

          (ii)  `D=0` , both roots are real and equal

         (iii) `D<0` ,  both roots are complex.

Given equation is   `6b^2+b+3=-5`

or,                     `6b^2+b+3+5=0`

or,                     `6b^2+b+8=0`

Here     `p=6, q=1, r=8.`

So,      `D=1^2-4.6.8=1-192=-191<0` .

So, by condition (iii) both roots of the given quadratic equation will be complex.

jeew-m eNotes educator| Certified Educator

`6b^2+b+3 = -5`

`6b^2+b+3+5 = 0`

`6b^2+b+8 = 0`


Discriminant = `(1)^2-4xx6xx8 = -191 < 0`

Since the discriminant is negative there are no real solutions to the above function. But there are two imaginary solutions for this using complex numbers.



oldnick | Student


adding 5 both sides:  `6b^2+b+3+5= -5+5`   `6b^2+b+8=0```
`Delta=1^2-4(8)(6)= 1-192= -191`

since `Delta <0`  the equation has two  complex coniugates solutions:

`b=(-1 +- sqrt(Delta))/12` `=(-1 +- i sqrt(191))/12`







pramodpandey | Student









`` Discriminant=-191 <0

so no real roots.Roots are complex conjugate.