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Question 38925:
xｳ + x - 2 = 0
Found 2 solutions by fractalier, AnlytcPhil:
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! We can see that x = 1 is a solution of
x^3 + x - 2 = 0
but if we factor that expression by x - 1, we get
(x - 1)(x^2 + x + 2) = 0
and the second polynomial is not factorable...you have to use the quadratic...
so x = 1 and
x = (-1 ｱ i*sqrt(7)) / 2
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website! xｳ + x - 2 = 0
The possible rational solutions have numerators which divide
evenly into the absolute value of the last term. The
possible denominators are those that divide evenly into
the absolute value of the coefficient of the highest power
of x.
Possible numerators: divisors of 2 which are: 1, 2
Possible denoiminators: divisors of 1, of which there is only one: 1
Possible rational solutions (they can be positive or negative): ｱ1, ｱ2
Try x = 1, using synthetic division
xｳ + x - 2 = 0
We must first insert a place-holder term for xｲ, namely 0xｲ
xｳ + 0xｲ + x - 2 = 0
1 | 1 0 1 -2
| 1 1 2
1 1 2 0
The last number on the bottom is zero, so
we are lucky to have found a solution on
the first trial.
Now we have factored the left side of
xｳ + x + 2 = 0
as
(x - 1)(xｲ + x + 2) = 0
Us the 0-factor principle:
x - 1 = 0 gives solution x = 1
xｲ + x + 2 = 0 must be solved by the
quadratic formula:
____________
-(1) ｱ ﾖ(1)ｲ-4(1)(2)
x = 覧覧覧覧覧覧覧覧覧覧覧
2(1)
______
-1 ｱ ﾖ1 - 8
x = 覧覧覧覧覧覧覧
2
__
-1 ｱ ﾖ-7
x = 覧覧覧覧覧
2
_
-1 ｱ iﾖ7
x = 覧覧覧覧覧
2
_
x = -1/2 ｱ ﾖ7/2ｷi
So the three solutions are
_ _
1, -1/2 + ﾖ7/2ｷi, and -1/2 - ﾖ7/2ｷi
Edwin
AnlytcPhil@aol.com
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