Equation help please?

Question

I need to show that for this equation:

3x^3 - 4x^2 + 5x - 2 = 0

there are no rational roots but it does have an irrational root between x = 0 and x=1.

thanks.

Answer 1

First, use Descartes' Rule of Signs:

f(x) = 3x^3 - 4x^2 + 5x - 2

This has three sign changes, so there are either three or one positive real root.

f(-x) = -3x^3 - 4x^2 - 5x - 2

This has no sign changes. It therefore has no negative real roots. Since the function is third order, it must have three roots. It therefore has three positive real roots and no complex ones or one positive real root and two complex ones. Either way, there's at least one positive real root.

The rational root theorem then tells you that the possible rational roots are 2, 1, 2/3 or 1/3. They are easy to check out - just plug them in and show that none of them are roots. Therefore, our one or three real roots must be irrational.

Lastly, check f(0) and f(1):

f(0) = -2
f(1) = 2

Since the sign of the function changes between 0 and 1, it must cross the x-axis somewhere in there. In other words, there's a root between 0 and 1. We know it isn't rational, so it must be irrational. QED

Answer 2

Well, just plug in 0 and 1 and see what you get. When you put 0 into the value of x, you get -2 = 0, and 1 into the value of x, you get 2 = 0. Therefore, ONLY a number BETWEEN 0 and 1 must exist for the value to come out to be an even 0.

Answer 3

call f (x ) = 3x^3 - 4x^2 + 5x -2

f (0) = -2
f (1) = 2

=> f (0) * f (1) < 0
=> there is at least 1 root between 0 and 1.

Answer 4

first factor from 3rd to 2nd order equation and then solve

(x)(3x^2 - 4x +5) = 2 or (3x^2 - 4x +5) = 2/x hope that helps.

you can use the quadratic formula on the left hand side of the equation since it is now only a second order expression.

Answer 5

3x^3 - 4x^2 + 5x = 2 ; x (3x^2 - 4x^2 + 5) = 2 ; x = 2/ 3x^2 - 4x +5