Find the discriminant of this quadratic equation then state the number and type of solutions. 6b^2+b+3= -5
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`
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.
`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.
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`
`` Discriminant=-191 <0
so no real roots.Roots are complex conjugate.