How would you try to solve \( x^3+x=4 \)?

So first I think about some concrete values. 0+0=0, 1+1=2, 8+2=10. OK, one solution between 1 & 2. No negative solution. Accessing calculus (which seems like cheating) derivative is \( 3x^2+1 \) so monotone increasing. I think that means this is a root with multiplicity 3, as one root, two imaginary is an 'S-curve' as my students say with the one root. (I should look into that at some point.)

Algebraically, I think about factoring as is, which seems like no help. If I'm right about \( (x-a)^3 \) for some \( a \)... that means \( x^3 - 3 a x^2+3 a^2 x-a^3=x^3+x-4 \). So is \( a=4^{1/3} \)? That doesn't work! And then there's no way for \( 3 a x^2 \) to be zero. So I take back what I said about multiplicity! (I really

*do*have to look at that more.)

So next I would look at numerically grinding it out. Something closer to 1 than 2 and proceed from there.

To solve it with a graphing calculator - piece of cake. At least for a decimal approximation. Here it is

on Desmos, along with Khayyam's geometric solution.

The last thing I want to do is to verify that his solution works. The question of how did he derive this is important, but I'm not going to get to it here.

So the equation of the circle is \( (x-b/2a)^2+y^2=(b/a)^2 \) and the parabola \( y=x^2/ \sqrt(a) \). So... hmm. Substituting the parabola equation into the circle gets us a quartic!

So why a circle? It gets us a right triangle, which gives us proportions.

So

\[ \frac{x}{(x^2/ \sqrt{a})} = \frac{x^2/ \sqrt {a}}{b/a - x} \\

\frac{\sqrt {a}}{x} = \frac{x^2/ \sqrt {a}}{b/a - x} \\

a(b/a-x) = x^3 \\

b - ax = x^3 \\

b = x^3 + ax \]

Sweet!

(The mathjax is displaying odd for me, \( \sqrt a \) is square root.)

I did think it was interesting that none of the students had any idea how Khayyam was drawing parabolas before graphing. Maybe we'll have to do some directrix learning.

p.s. Deborah Kent and Milan Sherman (Drake University) wrote a great extended piece on this.

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