Solution for October 11, 2006

$\alpha\,$ is a real root of $x^3 + 2x^2 + 10x - 20 = 0\,$ . Show that $\alpha ^2\,$ is irrational.

Easy Theory Solution by landen
By the rational roots theorem any rational root of this is an integer divisor of 20. We can check them all and none work.

Another possibility is to find the derivative of the cubic and notice it is always positive. So there is only 1 real root. Then we notice the polynomial is negative at $x=1\,$ and positive at $x=2\,$ , so the root is not an integer and is therefore irrational.

Now we have that $\alpha\,$ is irrational which is certainly required for the proposition to be true.

Assume $\alpha ^2 =c/d\,$ is rational with $c,d\,$ integers. So we now have that $\alpha\,$ is a root of $d\,x^2 - c.$

Now comes the major tactic. We use polynomial division to find that:

$x^3+2\,x^2+10\,x-20={{\left(x+2\right)\,\left(d\,x^2-c\right) }\over{d}}+{{\left(10\,d+c\right)\,x-20\,d+2\,c}\over{d}}$

If we substitute $x=\alpha\,$ we get:

$0 = 0 + {{\left(10\,d+c\right)\,\alpha-20\,d+2\,c}\over{d}}$

$\alpha={{20\,d-2\,c}\over{10\,d+c}}$

This gives that $\alpha\,$ is rational which is a contradiction so that the hypothesis that $\alpha ^2 \,$ is rational is false.
$\square\,$

Easy Theory Solution by Polytope
Establish that $\alpha\,$ is irrational as above. Next use polynomial division to get:

$x^4=\left(x-2\right)\,\left(x^3+2\,x^2+10\,x-20\right)-6\,x^2+40\,x -40$

If we substitute $x=\alpha\,$ we get:

$\alpha ^4= 0 -6\,\alpha ^2+40\,\alpha -40$

$\alpha ^4 +6\,\alpha ^2+40=40\,\alpha$

If $α 2$ were rational then the left side of the equation is rational and the right side is irrational, which would be a contradiction.
$\square\,$

Really Easy Theory Solution by landen
By the rational roots theorem any rational root of this is an integer divisor of 20. We can check them all and none work.

$\alpha^3+2\,\alpha^2+10\,\alpha-20=0$

$\alpha={{20-2\,\alpha^2}\over{\alpha^2+10}}$

This says $\alpha\,$ is rational if $\alpha ^2 \,$ is rational which is a contradiction. So $\alpha ^2 \,$ is irrational.
$\square\,$

Solution using CAS

The real root of $x^3 + 2x^2 + 10x - 20 =\,$
$-{{\left(3\,2^{{{1}\over{6}}}\,\sqrt{3}\,\sqrt{5}\,\sqrt{131}-88\,2 ^{{{2}\over{3}}}\right)\,\left(3\,\sqrt{2}\,\sqrt{3}\,\sqrt{5}\, \sqrt{131}+176\right)^{{{2}\over{3}}}-169\,2^{{{1}\over{3}}}\,\left( 3\,\sqrt{2}\,\sqrt{3}\,\sqrt{5}\,\sqrt{131}+176\right)^{{{1}\over{3 }}}+338}\over{507}}$

Then $\alpha ^2 =\,$
${{\left(3\,\sqrt{2}\,\sqrt{3}\,\sqrt{5}\,\sqrt{131}+176\right)^{{{1 }\over{3}}}\,\left(13\,2^{{{5}\over{6}}}\,\sqrt{3}\,\sqrt{5}\,\sqrt{ 131}-988\,2^{{{1}\over{3}}}\right)+\left(4\,2^{{{1}\over{6}}}\, \sqrt{3}\,\sqrt{5}\,\sqrt{131}-61\,2^{{{2}\over{3}}}\right)\,\left(3 \,\sqrt{2}\,\sqrt{3}\,\sqrt{5}\,\sqrt{131}+176\right)^{{{2}\over{3}} }-2704}\over{507}}$

Next we "guess" that $\alpha ^2\,$ solves the equation:

$x^3+16\,x^2+180\,x-400=0\,$

Polytope and landen used Maple and Maxima to show that $\alpha ^2\,$ is a root.

Next, $x^3+16\,x^2+180\,x-400=0\,$ has no rational roots by the rational roots theorem. Therefore, $\alpha ^2\,$ is irrational.
$\square\,$

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