Determining the number of non-real roots. Multiple choice strategy.

Question

I was tutoring a student, and this question came up on a mock standardized test. Assume the student has learned advanced precalculus at a high school level and has access to a graphing calculator. The question is meant to be solved in a few minutes, tops.

Which of the following equations has exactly two non-real solutions?

(A) $\quad x^3-x^2+2x-2 = 0$

(B) $\quad x^3 -2x^2 -5x +6 = 0$

(C) $\quad x^4-7x^2+12=0$

(D) $\quad x^3-8x^2+11x+20=0$

(E) $\quad x^4-5x^3+x^2 +25x-30=0$

The first strategy that came to my mind is using Descartes's Rule of Signs, but that test seems to be inconclusive. A graphical approach seems to be another way to approach it, but that seems inefficient to me, and not entirely reliable in a test taking scenario- where you may fumble with manipulating window sizes and whatnot.

The answer is cited as (A), and indeed, the first answer choice can be easily factored as $(x^2+2)(x-1)$, which verifies it as a solution (note: I haven't checked the roots of the other polynomials). But again, I do not expect that factoring should have been the intended approach here.

Does anyone see a quick method to solve this, or is this a poorly designed question?

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Edit: To followup on this question. Indeed, Gregory was correct in the comments. This question was from a unofficial SAT II Math practice exam. While the SAT II Math test does not require a calculator for any of its questions, it doesn't prohibit the use of one. It doesn't even ban computer algebra systems, like those on, say, the TI-89. Now, this particular practice test was constructed with (some) questions designed to train students to use the algebra software.

Thanks for the suggested solutions, though. I like the calculus methods, and I think my student will appreciate them as well.

Answer 1

If your test-taking strategy is savvy, you will try to eliminate the easier possibilities first; here, that means looking at the cubics first. Advanced pre-calc nowadays mentions how to take the derivative of a polynomial, so you can look at (A) and find that it has no real minimum or maximum, because the derivative $$ 3x^2-2x+2 $$ has a discriminant of $b^2-4ac = 4-24 < 0$. (That is, in trying to solve for $x$ using the quaadratic formula, you get no real solutions.) So (A) is monotone increasing, and can have only one point at which it crosses $y=0$, thus exactly two non-real solutions.

FInally, using the fact that this is a test question with a unique correct choice, you know the others all have zero (or 4) non-real solutions.

Answer 2

Hints for the quartics:

• (C) $\;x^4-7x^2+12=0$  This is a biquadratic i.e. a quadratic in $x^2\,$. By the discriminant and rule of signs it's easy to tell that it has two positive roots in $x^2\,$, therefore $4$ real roots in $x$.

• (E) $\;x^4-5x^3+x^2 +25x-30=x^4-5x^3+6x^2 - 5\cdot (x^2-5x+6)\,$.

Answer 3

If this were a MCQ with one or more than one correct answers, We'll need to check every option (You haven't specified explicitly that's why I'm taking full proof case)

Let $f$ be your cubic function. Just take derivative of each function. By the roots of $f'$ (Which is obviously a quadratic in case of cubic .. ). You will get the point of local maxima and minima of $f$.Let them be at $x=\alpha$ and $x=\beta$.

Now all you have to do is check sign of $f(\alpha)$ and $f(\beta)$ (Suppose $\alpha > \beta$ and leading coefficient to be positive.)

There are $4$ possibilities :

• $f(\alpha) >0$ ,$f(\beta) >0 \implies$ the function turns above $x$-axis and thus only one real root, rest two are complex.

• $f(\alpha) > 0$ ,$f(\beta) <0 \implies$ the function turns above $x$-axis and again below $x$-axis and thus only three real roots.

• $f(\alpha) <0$ ,$f(\beta) <0 \implies$ the function turns below $x$-axis and both the times thus only one real root, rest two are complex..

• $f(\alpha) <0$ ,$f(\beta) >0~$ Never possible.

• If $f'$ doesn't has real roots $\implies$ the cubic is monotonously increasing .Hence one real (repeated) root.

This method works well for cubic polynomials.

For bi-quadratic given in option $(C)$, it's actually a quadratic in $x^2$.

For biquadratic in option $(E)$, We'll need to factorize it as $(x^2-5x+6)(x^2-5)$ (as done by @dxiv in his answer).

Rest are all cubics.