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

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rakesh05's profile pic

Posted on

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's profile pic

Posted on

`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's profile pic

Posted on


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's profile pic

Posted on









`` Discriminant=-191 <0

so no real roots.Roots are complex conjugate.

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