I got to (2x-1)[(x-2)(x+1)] = 0 what next?
please show your steps
(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0?
2009-06-29 5:49 am
回答 (10)
2009-06-29 6:01 am
✔ 最佳答案
(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0(x-2)(x+1)[(x+1)+(x-2)]=0
(x-2)(x+1)(2x-1)=0
Therefore, three possible answers for x:
(2x-1) = 0, x = 1/2
(x+1) = 0, x = -1
(x-2) = 0, x = 2
X = 2, -1 or 1/2
Proof: Substitute these 3 numbers to the equation, will prove the equation correct
2009-06-29 7:15 am
(x - 2)(x + 1)^2 + (x + 1)(x - 2)^2 = 0
(x - 2)[(x + 1)^2 + (x + 1)(x - 2)] = 0
(x - 2)(x + 1)[(x + 1) + (x - 2)] = 0
(x - 2)(x + 1)(x + 1 + x - 2) = 0
(x - 2)(x + 1)(x + x + 1 - 2) = 0
(2x - 1)(x - 2)(x + 1) = 0
2x - 1 = 0
x = 1/2 (0.5)
x = 0
x - 2 = 0
x = 2
x + 1 = 0
x = -1
∴ x = -1, 1/2 (0.5), 2
(x - 2)[(x + 1)^2 + (x + 1)(x - 2)] = 0
(x - 2)(x + 1)[(x + 1) + (x - 2)] = 0
(x - 2)(x + 1)(x + 1 + x - 2) = 0
(x - 2)(x + 1)(x + x + 1 - 2) = 0
(2x - 1)(x - 2)(x + 1) = 0
2x - 1 = 0
x = 1/2 (0.5)
x = 0
x - 2 = 0
x = 2
x + 1 = 0
x = -1
∴ x = -1, 1/2 (0.5), 2
2009-06-29 6:09 am
This IS correct so far.
Now set each of the three factors equal to zero:
2x-1 = 0 and x-2 = 0 and x+1 = 0
Solve each equation for x:
x = 1/2 and x = 2 and x = -1.
Now set each of the three factors equal to zero:
2x-1 = 0 and x-2 = 0 and x+1 = 0
Solve each equation for x:
x = 1/2 and x = 2 and x = -1.
2009-06-29 6:02 am
(2x-1)(x-2)(x+1)=o
=>2x-1=o /OR/ x-2=0 /OR/ x+1=0
x=1/2
OR
x=2
OR
x=-1
=>2x-1=o /OR/ x-2=0 /OR/ x+1=0
x=1/2
OR
x=2
OR
x=-1
2009-06-29 6:02 am
2x-1 = 0
x= 1/2
x-2 = 0
x=2
x+1 = 0
x= (-1)
x= 1/2
x-2 = 0
x=2
x+1 = 0
x= (-1)
2009-06-29 6:00 am
(x-2)(x+1)[(x+1)+(x-2)]=0
x=2,x=-1 &
2x-1=0,x=1/2
x=2,x=-1 &
2x-1=0,x=1/2
2009-06-29 5:57 am
(2x-1)[(x-2)(x+1)] = 0
so 2x-1 = 0
or x-2 = 0
or x+ 1= 0
ie x =-1, 1/2 or 2 are solutions
check
(x-2)(x+1)[ x+1 + x-2] = 0
(x-2)(x+1)[2x -1]=0
so 2x-1 = 0
or x-2 = 0
or x+ 1= 0
ie x =-1, 1/2 or 2 are solutions
check
(x-2)(x+1)[ x+1 + x-2] = 0
(x-2)(x+1)[2x -1]=0
2009-06-29 5:56 am
what you have is not quite correct
the first thing to see is the two terms are the same.
the second thing to do is to expand it inside:
(x+1)(x-2) = x^2-x-2 (1)
then since there is two of the same term, (1) x 2
0 = 2x^2 - 2x - 4
quadratic equation with b = -2, c = -4, a = 2:
x = 0.5 +/- 0.25*sqrt(4-(4)(-4)(2))
x = 0.5 +/- 0.25*sqrt(36)
x = 0.5 +/- 0.25*6
x1 = 1/2 + 3/2 = 2
x2 = 1/2 - 3/2 = -1
the first thing to see is the two terms are the same.
the second thing to do is to expand it inside:
(x+1)(x-2) = x^2-x-2 (1)
then since there is two of the same term, (1) x 2
0 = 2x^2 - 2x - 4
quadratic equation with b = -2, c = -4, a = 2:
x = 0.5 +/- 0.25*sqrt(4-(4)(-4)(2))
x = 0.5 +/- 0.25*sqrt(36)
x = 0.5 +/- 0.25*6
x1 = 1/2 + 3/2 = 2
x2 = 1/2 - 3/2 = -1
2009-06-29 5:56 am
so
2x-1=0 , x= 1/2
x-2=0, x=2
x+1=0, x=-1
so x= 1/2,2 and -1
2x-1=0 , x= 1/2
x-2=0, x=2
x+1=0, x=-1
so x= 1/2,2 and -1
2009-06-29 6:04 am
(x-2)(x+1)^2 + (x+1)(x-2)^2 = 0
First we have to rewrite it like this since the items in parentheses are squared, we must write them twice:
(x-2)(x+1)(x+1) + (x+1)(x-2)(x-2) = 0
Now we have to FOIL these out:
(x² - x - 2)(x+1) + (x² - x - 2)(x-2) = 0
Go and multiply these out:
(x³ - x² + x² - x - 2x - 2) + (x³ - 2x² - x² - 2x - 2x + 4) = 0
Now combine like terms:
(x³ - 3x - 2) + (x³ - 3x² - 4x + 4) = 0
Get rid of the parentheses and combine like terms again:
2x³ - 3x² - 7x + 2 = 0
Now you can plug it into a calculator and graph it to find the zeros:
x = -1.4319 , .26151 , and 2.6704
First we have to rewrite it like this since the items in parentheses are squared, we must write them twice:
(x-2)(x+1)(x+1) + (x+1)(x-2)(x-2) = 0
Now we have to FOIL these out:
(x² - x - 2)(x+1) + (x² - x - 2)(x-2) = 0
Go and multiply these out:
(x³ - x² + x² - x - 2x - 2) + (x³ - 2x² - x² - 2x - 2x + 4) = 0
Now combine like terms:
(x³ - 3x - 2) + (x³ - 3x² - 4x + 4) = 0
Get rid of the parentheses and combine like terms again:
2x³ - 3x² - 7x + 2 = 0
Now you can plug it into a calculator and graph it to find the zeros:
x = -1.4319 , .26151 , and 2.6704
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