After referring to this question, I tried a similar approach to solve the below and got an incorrect answer.

Find the values of $p$ so that the equation $2\cos^2x - (p+3)\cos\ x + 2(p-1) = 0$ has real roots.

The solution to the question (given in the answer section of the book I referred) is as follows $-$

Solution to Question

The above solution is correct but, Why do I need to solve the equation for $\cos(x)$ and then find the values of $p$ instead of just applying $b^2-4ac \ge 0$ to the coefficients of the equations which are $2,\ (p+3)$ and $2(p-1)$ to get the values of $p$ such that the equation has real roots?

  • 3
    $\begingroup$ Your method would have worked for $2 x^2 - (p+3) x + 2(p-1) = 0$ as $x$ can take any real value. But it would not work for quadratic in $\cos x$ as there is an additional constraint of $|\cos x| \leq 1$ $\endgroup$
    – Math Lover
    Oct 8, 2021 at 12:39
  • $\begingroup$ In fact if this had just been an ordinary polynomial in a real variable and not a polynomial of $\cos x,$ you could set $p$ to any real number and you would always have two real solutions, $(p-1)/2$ and $2.$ $\endgroup$
    – David K
    Oct 8, 2021 at 12:44
  • $\begingroup$ So, Does it mean, because of the constraint of $0\le cos\ x \le 1$, directly applying $b^2-4ac\ge 0$ fails? $\endgroup$ Oct 8, 2021 at 12:49
  • $\begingroup$ @MathLover, I missed to mention you in the above comment. $\endgroup$ Oct 8, 2021 at 12:57
  • 1
    $\begingroup$ @AtheeshThirumalairajan No, re-read Math Lover's first comment. The $(b^2 - 4ac)$ analysis is necessary but not sufficient. You also have to consider that $-1 \leq \cos x \leq 1.$ Any value of $p$ that creates $1$ or $2$ real roots, none of which are in the interval $[-1,1]$ would have to be rejected. $\endgroup$ Oct 8, 2021 at 13:03

1 Answer 1


The problem to be solved was:

Find the values of $p$ so that the equation $2\cos^2 x - (p+3)\cos x + 2(p-1) = 0$ has real roots.

This is not the same as a problem like finding $p$ such that the equation $ 2 y^2 - (p+3)y + 2(p-1) = 0 $ has real roots, because we don't just have a polynomial in some variable $y$; the equation also involves $\cos x$.

When we ask for real roots we are asking for real numbers that can be plugged in for $x$ so as to make the equation $2\cos^2 x - (p+3)\cos x + 2(p-1) = 0$ true. Now since $x$ must be real, we know $\cos x$ also will be real (the cosine of any real number is a real number). More specifically, $\cos x$ will be a real number belonging to the closed interval $[-1,1].$ But it can be any number in that interval; if we let $x$ run from $0$ to $\pi$ then $\cos x$ will hit every number in $[-1,1].$

The thing is, if we have to start worrying about the "interval $[-1,1]$" part right away, it's hard to see how to solve the problem. So let's just worry about the "real number" part at first. We want to know that there is a real number that you can put in place of $\cos x$ in the equation that makes the equation true.

Or to put it slightly differently, let's define the variable $y$ by the relationship $y = \cos x,$ and now we want to know that there is a real number $y$ that satisfies $$ 2 y^2 - (p+3)y + 2(p-1) = 0 . $$

You already know how to do that: you must ensure that $b^2 - 4ac \geq 0.$ In this particular problem, $$b^2 - 4ac = (p + 3)^2 - 16(p-1) = p^2 - 10p + 25 = (p - 5)^2.$$

Now observe that $(p - 5)^2 \geq 0$, and therefore $b^2 - 4ac \geq 0,$ for any real number $p.$

That is, for any real number $p,$ there is at least one real number, usually two, that could be the value of $y$ in the true equation $ 2 y^2 - (p+3)y + 2(p-1) = 0. $

But now we have to remember that $\cos x$ cannot be just any real number. It has to be in the interval $[-1,1].$ So instead of just being concerned about whether a real number exists that makes $ 2 y^2 - (p+3)y + 2(p-1) = 0 $ true when we plug this number in for $y,$ we need to think about which real number we might have to plug in for $y,$ and make sure that that number is in the interval $[-1,1].$

The use of the variable $p$ makes it a little difficult to guess a factorization of the polynomial $2 y^2 - (p+3)y + 2(p-1),$ so let's just apply the standard formula to find solutions of any solvable quadratic equation in $y$:

$$ y = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}. $$

Putting $a = 2,$ $b = -(p+3),$ $c = 2(p-1),$ we have

$$ y = \frac{(p+3) \pm \sqrt{(p+3)^2 - 16(p-1)} }{4} = \frac{(p+3) \pm (p - 5)}{4}. $$

If we choose $+$ for $\pm$ we get $y = (p - 1)/2$ and if we choose $-$ we get $y = 2.$

Obviously $y = 2$ is no good because $2$ is not in the interval $[-1,1]$ so it cannot be a value of $\cos x.$ So our only hope is that the solution is $y = (p - 1)/2.$ In order for this to be a value of $\cos x$ (so that it solves the problem that was originally given) it must be true that $-1 \leq (p - 1)/2 \leq 1.$ The rest of the solution is just algebra to simplify these inequalities. The simplified version is $-1 \leq p \leq 3.$

Now it may seem that the book's solution did not actually start by requiring that $b^2 - 4ac \geq 0$ and asking what $p$ made that inequality true. You might ask what allows them to skip that step. The reason they can do it is because the test for $b^2 - 4ac \geq 0$ is itself just a consequence of the formula

$$ y = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}. $$

If the expression under the $\sqrt{\phantom{0}}$ symbol is negative then it does not have a real square root and the formula does not give you any solutions. So that expression has to be non-negative, that is, $b^2 - 4ac \geq 0.$ And if all you care about is whether there is any solution, not what particular number the solution is, $b^2 - 4ac \geq 0$ is all you need to know, because then there will be a real square root and you'll combine it with other real numbers to get a real number (or two) in the end.

(I'm assuming that you're working in real numbers only and not complex numbers. If you allow complex numbers then there is a non-real square root and we have to do some extra work to show that the final result of calculating $y$ is not real.)

Since we actually need to know more about $y$ than just that it's a real number, we need to finish evaluating the formula. In the book's solution they go directly to that step, but while simplifying $\sqrt{(p+3)^2 - 16(p-1)}$ to $\pm(p-5)$ they had to figure out that $(p+3)^2 - 16(p-1) = (p - 5)^2,$ which implies that $b^2 - 4ac \geq 0.$

  • $\begingroup$ +1 : very nice answer. $\endgroup$ Oct 10, 2021 at 6:59

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