# Irrational root of a function

DISCLAIMER: I apologize in advance if this question is naive. Every suggestion on how to approach the following problem will be very much appreciated.

I'm interested in the root of the following function:

$f(x)= [(2x-x^2)\cdot (2x-x^2)] -x$

If I use some software to draw this function, I can see a root around x=0.38, let's call it R.

my questions are: A) How can I calculate R? I don't know how to handle equations of 4th order B) I'm very interested in knowing if R is irrational.

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Expanding out your polynomial we have that $$f(x) = x^4-4x^3+4x^2-x.$$

0 is obviously a root. By the rational root test, the only possible other rational roots of the polynomial are $\pm 1$. Fortunately, $f(1) = 0$, so by dividing out by $x-1$ we can easily factor $$f(x) = x(x-1)(x^2-3x+1).$$

Using the quadratic formula we get all of $f$'s roots: $0, 1, \frac{3\pm \sqrt{5}}{2}.$

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+1. In particular, $\frac{3-\sqrt 5}{2} \approx 0.38$ is irrational. – Jonathan Christensen Jan 21 '13 at 16:38

Expanding the polynomial, we have:

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

Now we factor:

$$f(x) = x(x-1)(x^2-3 x+1)$$

Now either $x$, $x-1$ or $x^2-3 x+1$ is 0. So we have $x = 0, 1$. Now we solve the quadratic:

$$0 = x^2-3 x+1$$

$$x = \frac{1}{2} \left(3\pm\sqrt{5}\right)$$

Neither of which are rational. The difference of a rational and irrational is irrational, and the product of a rational and an irrational is also irrational. Since $\sqrt{5}$ is irrational, it follows that the roots of this equation are also irrational.

(More generally, if we have a polynomial of the second degree with integer coefficients, it follows that the solution is irrational iff $b^2 - 4ac$ is not a perfect square.)

Note that here, even though the equation is of the fourth degree, we can still factor it in order to reduce it to an easier situation composed of a quadratic and two linear factors.

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