x - 4x^2 + x^3 = 0
x = ?
How do you solve x - 4x^2 + x^3 = 0?
2011-11-07 10:30 am
回答 (7)
2011-11-07 11:19 am
x=0
x=2-sqrt(3)
x-2+sqrt(3)
sqrt > square root
press "Show Steps" on the link above
x=2-sqrt(3)
x-2+sqrt(3)
sqrt > square root
press "Show Steps" on the link above
2011-11-07 10:58 am
first factor out the x and equate it to zero,you will be left with 1-4x+x^2.you then use the quadratic equation to evaluate x....
2011-11-07 10:56 am
x - 4x^2 + x^3 = 0
=> x^3-4x^2+x=0
=> x(x^2-4x+1)=0
=> x=0 or (x^2-4x+1)=0
We have , x= [-b ± √(b^2-4ac)]/2a
here, b=-4, a=1 and c=1
Thus, x= (4± √16-4)/2
= (4± √12)/2
= (4± √4*3)/2
= (4±2√3)/2
= 2±√3
Sol.: x=0,2±√3
=> x^3-4x^2+x=0
=> x(x^2-4x+1)=0
=> x=0 or (x^2-4x+1)=0
We have , x= [-b ± √(b^2-4ac)]/2a
here, b=-4, a=1 and c=1
Thus, x= (4± √16-4)/2
= (4± √12)/2
= (4± √4*3)/2
= (4±2√3)/2
= 2±√3
Sol.: x=0,2±√3
2011-11-07 10:48 am
It cant solve by any method, you have to find its root so use Bisection Iterative method. Most easy and simple . . Bisection Method is like you will start by finding negative and postive no, it pretend as ur first root must exist between ur negative and positve no. Check dis out
f(x) = x-4x^2+x^3 = 0
x = 0
f(x) = 0 Postive no.
x=1
f(x) = -2 negative no
so root must exist between 0 and 1.
so now use formula of biseciton
x0 = 0, x1 = 1 root exist between
x2 = x0+x1/2
.
.
.
.
go on and on u will come to close to 0 not exact 0. Its numerical analysis problem not giving you exist 0, but near to 0.
f(x) = x-4x^2+x^3 = 0
x = 0
f(x) = 0 Postive no.
x=1
f(x) = -2 negative no
so root must exist between 0 and 1.
so now use formula of biseciton
x0 = 0, x1 = 1 root exist between
x2 = x0+x1/2
.
.
.
.
go on and on u will come to close to 0 not exact 0. Its numerical analysis problem not giving you exist 0, but near to 0.
參考: 2 years experience on solving numerical problems
2011-11-07 10:43 am
x^3 - 4x^2 + x = 0
x(x^2 - 4x + 1) = 0
one root is 0 (the lone x)
using the quadratic formula:
x = (4 +/- SQR(16 - 4(1)(1)))/2(1)
x = (4 +/- SQR(12))/2
x = (4 +/- 2SQR(3))/2
x ~ 0.268, 3.732
The 3 exact roots for x are 0, (4 - 2SQR(3))/2, (4 + 2SQR(3))/2
- .--
x(x^2 - 4x + 1) = 0
one root is 0 (the lone x)
using the quadratic formula:
x = (4 +/- SQR(16 - 4(1)(1)))/2(1)
x = (4 +/- SQR(12))/2
x = (4 +/- 2SQR(3))/2
x ~ 0.268, 3.732
The 3 exact roots for x are 0, (4 - 2SQR(3))/2, (4 + 2SQR(3))/2
- .--
2011-11-07 10:40 am
take x out
x(x^2-4x+1)
then use quadratic equation to solve it
x(x^2-4x+1)
then use quadratic equation to solve it
2011-11-07 10:38 am
x - 4x² + x³ = 0
x(1 - 4x + x²) = 0
x(x-2+√3)(x-2-√3) = 0
x=0 and x=2±√3
x(1 - 4x + x²) = 0
x(x-2+√3)(x-2-√3) = 0
x=0 and x=2±√3
參考: ,
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