# When the Rational Root Theorem Fails

On a math test, one of the questions was to solve $-\sqrt{7-x}=-\frac{x^2}{2}+12x-10$. I solved graphically with a calculator, but later tried to solve algebraically, when I had more time. The equation is equivalent (with extraneous solutions) to $x^4 + 48x^3 +536x^2 -956x + 372=0$. This quartic has no rational roots (the rational roots theorem gives ±1, ±2, ±3, ±4, ±6, ±12, ±31, ±62, ±93, ±124, ±186, ±372, but none of these are zeros. I have shown that all real zeros lie in (-62, 2), but this is as far as I have gotten. What is the next step after the Rational Roots Theorem (RRT) has failed to find a root?

Thanks!

The quartic formula! Didn't you learn it in school? :p

Hurkyl
Staff Emeritus
Gold Member
For many purposes, the best way to represent a root in this situation is something of the form:

r is the root of $x^4 + 48x^3 +536x^2 -956x + 372=0$ between 0.6 and 0.7​

Incidentally, I plugged your original equation and your quartic equation into wolframalpha and the solutions don't agree -- you've made an error in arithmetic or in transcription. (although correcting your arithmetic won't make things any easier)

The original equation, $-\sqrt{7-x} = -\frac{x^2}{2} + 12x - 10$ has two solutions, at x1≈0.996 and x2≈-25.242, but the quartic has two extra solutions: at x3≈0.607 and at x4≈-24.361. These last two are extraneous solutions, but x1 and x2 are also solutions of the quartic.

uart
The original equation, $-\sqrt{7-x} = -\frac{x^2}{2} + 12x - 10$ has two solutions, at x1≈0.996 and x2≈-25.242, but the quartic has two extra solutions: at x3≈0.607 and at x4≈-24.361. These last two are extraneous solutions, but x1 and x2 are also solutions of the quartic.

I'm pretty sure that the original equation has just one real solution. Write it as:
$$\sqrt{7-x} = \frac{x^2}{2} - 12x + 10$$
and it's clear that the LHS is the upper half of a sideways parabola, with domain $x \leq 7$, while the RHS is a vertical parabola with axis of symmetry $x=12$. Therefore, where the domains overlap both curves are monotonic. So there is only one real solution.

BTW. Fixed point iteration of $x = \frac{1}{12} (0.5*x^2 + 10 - \sqrt{7-x} )$ gives this solution at about 0.640259457069546.

I'm pretty sure that the original equation has just one real solution. Write it as:
$$\sqrt{7-x} = \frac{x^2}{2} - 12x + 10$$
and it's clear that the LHS is the upper half of a sideways parabola, with domain $x \leq 7$, while the RHS is a vertical parabola with axis of symmetry $x=12$. Therefore, where the domains overlap both curves are monotonic. So there is only one real solution.

BTW. Fixed point iteration of $x = \frac{1}{12} (0.5*x^2 + 10 - \sqrt{7-x} )$ gives this solution at about 0.640259457069546.

Your comment about one solution threw me off. We had gone over the problem (graphically, of course) in class, and there were two solutions. I then noticed that I had typed the wrong equation, and simply copied it from my first post. I switched the signs:

$$\sqrt{7-x} = \frac{x^2}{2} + 12x - 10$$

should be correct. Wolfram Alpha confirms that there are two zeros here. Regardless of the signs, how would I find these solutions algebraically, without relying on the quartic formula?

Thanks!

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uart
Your comment about one solution through me off. We had gone over the problem (graphically, of course) in class, and there were two solutions. I then noticed that I had typed the wrong equation, and simply copied it from my first post. I switched the signs:

$$\sqrt{7-x} = \frac{x^2}{2} + 12x - 10$$

should be correct. Wolfram Alpha confirms that there are two zeros here. Regardless of the signs, how would I find these solutions algebraically, without relying on the quartic formula?

Thanks!

Yes obviously that changes things. The axis of symmetry of that parabola is x = -12, so that it is now possible for the $\sqrt{7-x}$ curve to intersect with both "sides" of the parabola (I haven't checked that it actually does, but will take your word for it). With the previously given equation however, this was simply not possible. 