Solve for X^3 -x=1/4(x^2 -1) Thanks?
回答 (4)
2016-07-20 10:03 am
✔ 最佳答案
Hello,x³ - x = (1/4)(x² - 1)
let's multiply both sides by 4:
4 (x³ - x) = 4 (1/4)(x² - 1)
4(x³ - x) = x² - 1
let's collect all terms at the left side:
4(x³ - x) - (x² - 1) = 0
let's factor x out of the first binomial, yielding the common factor (x² - 1):
4x (x² - 1) - (x² - 1) = 0
let's factor out (x² - 1):
(x² - 1) (4x - 1) = 0
let's further factor x² - 1 (as a difference of squares):
(x² - 1²) (4x - 1) = 0
(x + 1)(x - 1)(4x - 1) = 0
finally let's equate each factor to zero:
x + 1 = 0 → x = - 1
x - 1 = 0 → x = 1
4x - 1 = 0 → 4x = 1 → x = 1/4
thus the solutions are:
x₁ = - 1
x₂ = 1/4
x₃ = 1
I hope it helps
2016-07-20 9:58 am
x³ - x = (1/4)(x² -1)
4(x³ - x) = x² -1
4(x³ - x) - (x² -1) = 0
4x(x² - 1) - (x² -1) = 0
(4x - 1)(x² - 1) = 0
(4x - 1)(x - 1)(x + 1) = 0
x = 1/4 or x = 1 or x = -1
4(x³ - x) = x² -1
4(x³ - x) - (x² -1) = 0
4x(x² - 1) - (x² -1) = 0
(4x - 1)(x² - 1) = 0
(4x - 1)(x - 1)(x + 1) = 0
x = 1/4 or x = 1 or x = -1
2016-07-20 10:20 am
x³ - x = (1/4).(x² - 1)
x.(x² - 1) = (1/4).(x² - 1)
[x.(x² - 1)] - [(1/4).(x² - 1)] = 0
(x² - 1).[x - (1/4)] = 0
(x + 1).(x - 1).[x - (1/4)] = 0
First case: (x + 1) = 0 → x + 1 = 0 → x = - 1
Second case: (x - 1) = 0 → x - 1 = 0 → x = 1
Third case: [x - 1/4)] = 0→ x - (1/4) = 0 → x = 1/4
x.(x² - 1) = (1/4).(x² - 1)
[x.(x² - 1)] - [(1/4).(x² - 1)] = 0
(x² - 1).[x - (1/4)] = 0
(x + 1).(x - 1).[x - (1/4)] = 0
First case: (x + 1) = 0 → x + 1 = 0 → x = - 1
Second case: (x - 1) = 0 → x - 1 = 0 → x = 1
Third case: [x - 1/4)] = 0→ x - (1/4) = 0 → x = 1/4
2016-07-20 2:04 pm
X^3 -X=1/4[X^2-1]
X^3-X=2^-2[X.X/X]
X^3-X=2^-2.X
4X^3-4X-X=0
4X^2=5
X^2=5/4
X=1.118 Answer
X^3-X=2^-2[X.X/X]
X^3-X=2^-2.X
4X^3-4X-X=0
4X^2=5
X^2=5/4
X=1.118 Answer
2016-07-20 11:58 am
4x^3-x^2-4x+1 = 0
(x-1)(x+1)(4x-1) = 0
x = -1, 1/4, 1
(x-1)(x+1)(4x-1) = 0
x = -1, 1/4, 1
2016-07-20 9:59 am
x^3 – x = (1/4)(x^2 - 1) ………………..(1)
(x^3 – x)/(x^2 - 1) = x = 1/4
That is equivalent to (4x – 1) being a factor. Now rearrange (1) as
4x^3 – x^2 – 4x + 1 = (4x – 1)(x^2 + kx – 1)
Coefficient of x: -4 = -4 + k, so k = 0
(4x – 1)(x^2 – 1) = (4x – 1)(x – 1)(x + 1) = 0
Roots are 1/4, 1 and -1
(x^3 – x)/(x^2 - 1) = x = 1/4
That is equivalent to (4x – 1) being a factor. Now rearrange (1) as
4x^3 – x^2 – 4x + 1 = (4x – 1)(x^2 + kx – 1)
Coefficient of x: -4 = -4 + k, so k = 0
(4x – 1)(x^2 – 1) = (4x – 1)(x – 1)(x + 1) = 0
Roots are 1/4, 1 and -1
2016-07-20 9:57 am
x³ - x = (1/4)(x² - 1)
so, 4x³ - 4x = x² - 1
=> 4x³ - x² - 4x + 1 = 0
Now, x = 1 is a solution, so (x - 1) is a linear factor of 4x³ - x² - 4x + 1
i.e. 4x³ - x² - 4x + 1 = (x - 1)(4x² + bx - 1)
Equating coefficients gives b = 3
Hence, 4x³ - x² - 4x + 1 = (x - 1)(4x² + 3x - 1)
Also, (x - 1)(4x² + 3x - 1) = (x - 1)(4x - 1)(x + 1)
Therefore, x = 1/4 and x = -1 are also solutions
So, x = ±1 and 1/4 are the three solutions to 4x³ - x² - 4x + 1 = 0
:)>
so, 4x³ - 4x = x² - 1
=> 4x³ - x² - 4x + 1 = 0
Now, x = 1 is a solution, so (x - 1) is a linear factor of 4x³ - x² - 4x + 1
i.e. 4x³ - x² - 4x + 1 = (x - 1)(4x² + bx - 1)
Equating coefficients gives b = 3
Hence, 4x³ - x² - 4x + 1 = (x - 1)(4x² + 3x - 1)
Also, (x - 1)(4x² + 3x - 1) = (x - 1)(4x - 1)(x + 1)
Therefore, x = 1/4 and x = -1 are also solutions
So, x = ±1 and 1/4 are the three solutions to 4x³ - x² - 4x + 1 = 0
:)>
2016-07-20 9:46 am
x=1, x-1=0 is a root of above eqautaion .. divide the equation by this and u will get other two roots.
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