Thursday

August 28, 2014

August 28, 2014

Posted by **kat** on Sunday, May 13, 2007 at 2:55pm.

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

Hi kat,

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)

and

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[224]

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[224]

B = 15/8 - 1/8 sqrt[224]

And thus:

x = a + b =

(15/8 + 1/8 sqrt[224])^(1/3) +

(15/8 - 1/8 sqrt[224] )^(1/3) =

1/2 [(15 + sqrt[224])^(1/3) +

(15 - sqrt[224])^(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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