How do you determine if a number is rational or irrational?

[temaire]

Homework Statement

$$x^3-4x+4$$

Homework Equations

None

The Attempt at a Solution

I tried using the quadratic formula but it doesn't work. I started with this, but I don't know what to do now.$$x^{3}+0x^2-4x+4$$

 

[gabbagabbahey]

What exactly are you trying to do?!

$x^3-4x+4$ is an expression, not a problem statement! Do you mean that you are trying to find the roots of the equation $x^3-4x+4=0$?

 

[temaire]

yes I am trying to find the roots of the equation

 

[gabbagabbahey]

Well, there are no simple solutions to this cubic equation. You can try using the cubic formula found here.

Is this the entire question, or is this part of a larger problem? Are you sure you have have written the equation correctly?

 

[temaire]

This is the correct equation, and it is the entire question.

 

[gabbagabbahey]

Are you asked to use any specific method? (such as approximating the real root using Newton's method)
Are you allowed to use a calculator/computer?

 

[temaire]

No specific method is being asked but you can use a simple scientific calculator.

 

[gabbagabbahey]

What methods have you been taught in whichever course this is for? Use one of those.

 

[TayTayDatDude]

Edit~

 

[gabbagabbahey]

Try (x^3+0x^2-4x+4) = 0 (Just noticed you have)

After that, try "hunting" to find which number works. After you have found a number, use synthetic division.

There are no rational roots to this equation, so that method won't work.

 

[Дьявол]

Find the roots using Cardano's method.

$$D=(\frac{4}{2})^2 + (\frac{-4}{3})^3=4 - \frac{64}{27}=\frac{108-64}{27}=\frac{44}{27} > 0$$, so it got one rational and two complex roots.

The rational one is ~(-2.38)

 

[gabbagabbahey]

Find the roots using Cardano's method.

$$D=(\frac{4}{2})^2 + (\frac{-4}{3})^3=4 - \frac{64}{27}=\frac{108-64}{27}=\frac{44}{27} > 0$$, so it got one rational and two complex roots.

The rational one is ~(-2.38)

The root (-2.38) is real not rational. The rational roots theorem tells you that the only possible candidates for rational roots of this equation are $\pm 1$, $\pm 2$, and $\pm 4$...and since none of these are roots; there are no rational roots.

 

[Дьявол]

Rational numbers are also real numbers. So I suppose -2.38...(infinite number of decimals) is irrational number (also real number).

 

[gabbagabbahey]

Rational numbers are also real numbers. So I suppose -2.38...(infinite number of decimals) is irrational number (also real number).

A rational number is a real number that can be expressed as a ratio of two integers...for example, 1/3, 0.25, and 3658793659/548709568017097 are rational numbers.

However, things like $\pi$, $\sqrt{2}$ and any non-repeating non-terminating decimal are irrational numbers.