solve 4x^3-6x+2=0 for all x
2x^3 -3x +1 = 0
x = 1 is one solution. That is obvious by inspection. Therefore (x-1) is a factor of the equation.
Divide the cubic by (x-1) to get the quadratic factor, (2x^2 +2x -1). That can be solved using the quadratic equation to get the other two roots
x = (1/4)[-2 +/- sqrt 12]
= -(1/2) +/- (1/2)sqrt3
by the way, this is algebra, not aclculus
this is part of calculus cause i need it to find critical points of a graph
To solve the equation 4x^3 - 6x + 2 = 0 for all x, we can use a combination of algebraic techniques, including factoring and the use of the quadratic formula.
Step 1: Factor out the greatest common factor (GCF)
In this case, there is no common factor other than 1, so we can skip this step.
Step 2: Determine if the equation can be factored
In this equation, it is not immediately apparent how to factor it, since all terms have different coefficients. So, we'll move on to the next step.
Step 3: Apply the Rational Root Theorem (if necessary)
The Rational Root Theorem can help us determine possible rational roots of the equation. However, in this case, it is not applicable since the Rational Root Theorem only works for equations with integer coefficients, and our equation contains decimal coefficients.
Step 4: Use the quadratic formula
Since the equation is a cubic equation (highest degree term is x^3), we can solve it by using the quadratic formula for a depressed cubic equation. The depressed cubic equation is obtained by making a substitution.
The substitution x = (y - b) / (3a) brings the equation to a depressed cubic:
y^3 + py + q = 0
where p = (3ac - b^2) / (3a^2) and q = (2b^3 - 9abc + 27a^2d) / (27a^3)
In our case, a = 4, b = 0, c = -6, and d = 2. Performing the substitution, we get:
y^3 - 8y - 16 = 0
Now, we can apply the quadratic formula to solve for y:
y = ( -q/2 + (q^2/4 + p^3/27)^0.5 )^(1/3) + ( -q/2 - (q^2/4 + p^3/27)^0.5 )^(1/3)
Step 5: Evaluate the cubic roots to obtain the values of y
Using a calculator or numerical methods, find the values of y. In this case, solving for y gives the following approximate solutions:
y ≈ 2.35
y ≈ -2.58
y ≈ 1.23
Step 6: Back-substitute to find the values of x
Now that we have y, we can back-substitute it into the equation x = (y - b) / (3a) to find the values of x:
For y ≈ 2.35: x ≈ (2.35 - 0) / (3 * 4) ≈ 0.156
For y ≈ -2.58: x ≈ (-2.58 - 0) / (3 * 4) ≈ -0.540
For y ≈ 1.23: x ≈ (1.23 - 0) / (3 * 4) ≈ 0.102
Therefore, the solutions to the equation 4x^3 - 6x + 2 = 0 for all x are approximately:
x ≈ 0.156
x ≈ -0.540
x ≈ 0.102