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.