quadratic formula

unique · 2006-10-12 04:58:08

use the quadratic formula to find the roots of the equation
3x^3 - 2x + 5 = 0

i did

formula :
-b +- sqrt (b^2 - 4ac)
x = ----------------------------
2a

- (-2) +- sqrt( (-2)^2 - 4(3)(5) )
x = ----------------------------------------
2(3)

2 +- sqrt (-4 - 4(15) )
x = ------------------------------
6

2 +- sqrt ( -4 - 60)
x = ------------------------------
6

2 +- sqrt ( -64 )
x = ---------------------------
6

now what i do i forgot?

Prakash Panneer · 2006-10-12 05:39:14

Is this a quadratic equaion?

Last edited by Prakash Panneer (2006-10-12 05:50:14)

unique · 2006-10-12 05:40:53

yes

Prakash Panneer · 2006-10-12 05:50:40

If it is a quadraic equation, then the equation is 3x^2-2x+5 = 0.

Then the solution is,

x = 2 +-Sqrt((-2)^2- 4*3*5)
____________________
2*3

x = 2 +- Sqrt(4 - 60)
_____________
6

x = 2+- Sqrt(-54)
____________
6

x = 2+- Sqrt( -9 * 6)
______________
6

x = 2 +- 3Sqrt(6) i
____________
6

x = 2 + 3 Sqrt(6) i and 2 - 3Sqrt(6) i
____________ ___________ ( Sqrt(-1) = i )
6 6

This equation has complex roots.

mathsyperson · 2006-10-12 05:59:16

Sorry Prakash, there's a mistake in there. 4-60 = -56, not -54.

x = 2 +- Sqrt(4 - 60)
_____________
6

x = 2+- Sqrt(-56)
____________
6

x = 2+- Sqrt( -4 * 14)
______________
6

x = 2 +- 2Sqrt(14) i
____________
6

x = 1 + Sqrt(14) i and 1 - Sqrt(14) i
____________ ___________ ( Sqrt(-1) = i )
3 3

If you haven't started learning about complex numbers yet, you can just say that there are no solutions.

Ricky · 2006-10-12 06:37:44

If you haven't started learning about complex numbers yet, you can just say that there are no solutions.

No real solutions.

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