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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?
Desi
Raat Key Rani !
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Is this a quadratic equaion?
Last edited by Prakash Panneer (2006-10-12 05:50:14)
Letter, number, arts and science
of living kinds, both are the eyes.
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yes
Desi
Raat Key Rani !
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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
____________
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x = 2 + 3 Sqrt(6) i and 2 - 3Sqrt(6) i
____________ ___________ ( Sqrt(-1) = i )
6 6
This equation has complex roots.
Letter, number, arts and science
of living kinds, both are the eyes.
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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.
Why did the vector cross the road?
It wanted to be normal.
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If you haven't started learning about complex numbers yet, you can just say that there are no solutions.
No real solutions.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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