quadratic formula

[Mickey]

In using the quadratic formula, what if the expression under the square root sign is negative. How do we compute this?

 

[galactus]

Then it is complex.

 

[Mickey]

Eg. here's the expression f(x) = 3x^2 - 2x + 4 = 0 and I have to find what x could be to make the expression true.

I tried x(3x-2) + 4 = 0 but I don't know where to go from here.

 

[galactus]

It has two complex solutions, so I would use the quadratic formula instead of factoring.

 

[Denis]

Forget that...what did you get using the quadratic ?

 

[BigGlenntheHeavy]

\(\displaystyle Discriminant \ of \ Quadratic:\)

\(\displaystyle b^{2}-4ac \ > \ 0, \ 2 \ real \ solutions\)

\(\displaystyle b^{2}-4ac \ = \ 0, \ 1 \ real \ solution\)

\(\displaystyle b^{2}-4ac \ < \ 0, \ 2 \ imaginary \ solutions\)

\(\displaystyle f(x) \ = \ 3x^{2}-2x+4, \ hence \ x \ = \ \frac{[2 \pm(4-48)^{1/2}]}{6}\)

\(\displaystyle = \ \frac{[2 \pm(4(-11))^{1/2}]}{6} \ = \ \frac{[2 \pm2i\sqrt11]}{6} \ = \ \frac{[1 \pm(i) \sqrt11]}{3}, \ i \ = \ \sqrt-1.\)

 

[Mickey]

but the quadratic formula gives you


x = -(-2)^2 +/- [(-2)^2 - 4(3)(4)}^1/2 divided by 2(3)

= 4 +/- (-44)^1/2 divided by 6

and then it bogs down since (-44)^1/2 is messy. But I still need to set 3x^2 -2x +4 = 0 and get values for x that satisfy this.

 

[Deleted member 4993]

but the quadratic formula gives you


x = -(-2)^2 +/- [(-2)^2 - 4(3)(4)}^1/2 divided by 2(3)

= 4 +/- (-44)^1/2 divided by 6

and then it bogs down since (-44)^1/2 is messy. But I still need to set 3x^2 -2x +4 = 0 and get values for x that satisfy this.

You are not paying attention to the answers - you have been told three times now

There is no real solution for x.

 

[fasteddie65]

By the way, the word is "disciminant" not "discriminate" for b^2 - 4ac.

 

[BigGlenntheHeavy]

The word is discriminant, not disciminant nor discriminate, thank you, I'll remember that.