Solve for x x^3-4x+2=0

Problem

x3−4*x+2=0

Solution

1. Identify the type of equation. This is a cubic equation of the form a*x3+b*x2+c*x+d=0 where a=1 b=0 c=−4 and d=2

2. Test for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient: ±1,±2

3. Evaluate the function ƒ(x)=x3−4*x+2 at these values.

ƒ(1)=1−4+2=−1

ƒ*(−1)=−1+4+2=5

ƒ(2)=8−8+2=2

ƒ*(−2)=−8+8+2=2

Since none of these are zero, there are no rational roots.

4. Apply the trigonometric method for cubic equations. For a cubic x3+p*x+q=0 we check the discriminant Δ=(q/2)2+(p/3)3

p=−4

q=2

Δ=(2/2)2+((−4)/3)3=1−64/27=−37/27

5. Determine the nature of the roots. Since Δ<0 there are three distinct real roots. We use the formula (x_k)=2√(,−p/3)*cos(1/3*arccos((3*q)/(2*p)√(,−3/p))−(2*π*k)/3) for k=0,1,2

6. Calculate the constants for the trigonometric substitution.

√(,−p/3)=√(,4/3)=2/√(,3)

cos(3*θ)=−q/2√(,−27/(p3))=−2/2√(,27/64)=−(3√(,3))/8

3*θ=arccos(−(3√(,3))/8)

7. Solve for the three values of x

(x_1)=4/√(,3)*cos(1/3*arccos(−(3√(,3))/8))

(x_2)=4/√(,3)*cos(1/3*arccos(−(3√(,3))/8)−(2*π)/3)

(x_3)=4/√(,3)*cos(1/3*arccos(−(3√(,3))/8)−(4*π)/3)

8. Approximate the values numerically.

(x_1)≈1.6751

(x_2)≈0.5392

(x_3)≈−2.2143

Final Answer

x=4/√(,3)*cos(1/3*arccos(−(3√(,3))/8)−(2*π*k)/3),k∈{0,1,2}

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