Posted by kat on Sunday, May 13, 2007 at 2:55pm.
which solution is the correct for this equation
4x^3 - 3x = 15
-1.8, -1.7, -1.6, -1.5, 1.5, 1.6, 1.7, 1.8 ??
What is your thinking, and why?
Have you considered graphing
y= 4x^3 -3x -15 ?
Kat, what level of study are you at.
This is an equation which can be "solved" only by sophisticated methods.
Are you supposed to use a programmable calculator?
I just used an iteration method called "Newton's Method" with a standard calculator and go an appr value of 1.7140 in 5 iterations for your equation.
This question falls well within the Calculus and post-Calculus area of study.
its university level
Let me show you how you can solve this equation exactly. You write the equation as:
4x^3 = 3x + 15 (1)
The trick is to compare this with the equation for (a+b)^3:
(a+b)^3 = a^3 + 3 a^2b + 3 a b^2 + b^3
You can rewrite this as:
(a+b)^3 = 3ab(a+b)+ a^3 + b^3
Multiply both sides by 4:
4(a+b)^3 = 12ab(a+b)+ 4(a^3 + b^3) (2)
If you compare equation (1) to equation(2) you see that if you can somehow choose a and b such that:
12ab = 3 (3)
4(a^3 + b^3) = 15 (4)
then a solution of eq. (1) will be
x = a + b
So, our next task is to solve eq. (3) and
eq. (4) for a and b. Let's take the third power of eq. (3):
a^3 b^3 = 1/64
If we now put:
A = a^3
B = b^3
then we have:
A B = 1/64
and, from eq. (4):
A + B = 15/4
Solving the two equations for A and B is easy, you get a quadratic equation:
A (15/4 - A) = 1/64 ---->
A^2 -15/4 A + 1/64 = 0 --->
A = 15/8 +/- 1/8 sqrt
We don't need to calculate B, because the symmetry with respect to interchanging A and B implies that solutions for B are also solutions for A. You can thus take the upper sign to be A and the lower sign to be B (or vice versa):
A = 15/8 + 1/8 sqrt
B = 15/8 - 1/8 sqrt
x = a + b =
(15/8 + 1/8 sqrt)^(1/3) +
(15/8 - 1/8 sqrt )^(1/3) =
1/2 [(15 + sqrt)^(1/3) +
(15 - sqrt)^(1/3)]
which is about 1.71401459.. (also given by Reiny above).
Third degree equtions have 3 solutions, so you may not always find the desired solution this way. To find the other two you must use the fact that an equation of the form x^3 = a has three solutions:
x = a^(1/3) Exp[2 pi i n/3]
where n is an integer.
So, when we extracted the cube root of A and B to find a and b we could have multipled each term by Exp[2 pi i n/3]. We must still satisfy Eq. (3):
12ab = 3
so, the product of a and b must not change which means that if we multiply a by Exp[2 pi i n/3] we must multiply b by Exp[-2 pi i n/3]. This way you find three solutions for n = 0, 1 and 2.
This method can be used to solve any third degree equation of the form:
a x^3 + b x^2 + c x + d = 0,
as a simple substitution:
x = y - b/(3a)
will remove the quadratic term.
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