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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?

My approach :-

f(x)= $x^3-x^2-3x-9$

Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root

f(-x)= $-x^3-x^2+3x-9$

2 sign changes here, indicating at most 2 negative real roots

I end up with following 2 possibilities:-

1)1 positive, 2 negative real roots

2)1 positive, 2 imaginary roots

how to progress further ?

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  • $\begingroup$ For what it's worth, I posted a comment that suggested exploring the first and second derivatives of $f(x)$. Then, I noticed the algebra-precalculus tag in your posting. Consequently, I deleted my original comment, since derivatives are not part of the precalculus curriculum. $\endgroup$ Nov 4, 2021 at 7:43
  • $\begingroup$ @user2661923 , you can post the solution, there's no harm in learning different approaches :) $\endgroup$
    – Fin27
    Nov 4, 2021 at 8:26
  • $\begingroup$ ...except the two roots other than $ \ 3 \ $ are complex $ \ ( -1 \ \pm \ i·\sqrt2 ) \ \ , \ $ not purely imaginary. $\endgroup$
    – user882145
    Dec 9, 2022 at 6:07

5 Answers 5

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$$x^3-x^2-3x-9=(x-3)(x^2+2x+3)$$

Now you just have to compute the discriminant of the quadratic term to conclude that the roots are imaginary.

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Given the posted question's precalculus tag, I normally would not have posted the following answer. However, since Siong Thye Goh has given a precalculus answer, I will give an answer that involves derivatives, which is a concept from Calculus (AKA Real Analysis). Note that I am generally ignorant of the precalculus tools available to attack such a problem. Therefore, there may be other precalculus methods of attack besides that given in the answer by Siong Thye Goh.


$f(x)= x^3-x^2-3x-9.$
How many negative real roots does the equation $f(x) = 0$ have ?

The posted question is equivalent to asking how many times $f(x)$ crosses the $x$-axis, for $-\infty < x < 0.$ The behavior of $f(x)$ may be analyzed by analyzing its first and second derivatives, and by noticing that $f(x)$ is a continuous function.

Without belaboring the underlying theorems, what this implies is that if there is an interval $[a,b]$ such that $a < b$, and if there exists $x_1, x_2$ both in $[a,b]$ such that $f(x_1) < 0 < f(x_2)$, then there must exist at least one value $x_0$ in $[a,b]$ such that $f(x_0) = 0$. Note that it doesn't matter whether $x_1 < x_2$ or $x_1 > x_2$.

$f'(x) = 3x^2 - 2x - 3$ and $f''(x) = 6x - 2$.

$\displaystyle f'(x) = 0 \implies ~x ~= ~\left(\frac{1}{6}\right) ~\left[ ~2 ~\pm ~\sqrt{4 + 36} ~\right] ~= ~\left(\frac{1}{3}\right) ~\left[ ~1 ~\pm ~\sqrt{10} ~\right].$

Since the posted question is concerned only about negative roots, attention can be confined to
$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right) ~\left[ ~1 ~- ~\sqrt{10} ~\right].$

Since $f''(x) < 0$, for all $x < 0$, $f(x)$ is maximized at $x = x_3$, for all $x < 0$.

$\displaystyle (x_3)^3 ~= ~\left(\frac{1}{27}\right) ~\left[ ~1 - 3\sqrt{10} + 30 - 10\sqrt{10} ~\right] ~= ~\left(\frac{1}{27}\right) ~\left[ 31 - 13\sqrt{10} ~\right].$

$\displaystyle (x_3)^2 ~= ~\left(\frac{1}{9}\right) ~\left[ ~1 - 2\sqrt{10} + 10 ~\right] ~= ~\left(\frac{1}{27}\right) ~\left[ 33 - 6\sqrt{10} ~\right].$

$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right) ~\left[ ~1 - \sqrt{10} ~\right] ~= ~\left(\frac{1}{27}\right) ~\left[ 9 - 9\sqrt{10} ~\right].$

Therefore,

$$f(x_3) = \left(\frac{1}{27}\right) \times ~\left\{ ~\left[ 31 - 13\sqrt{10} ~\right] ~- ~\left[ 33 - 6\sqrt{10} ~\right] ~- ~\left[ 27 - 27\sqrt{10} ~\right] ~- ~\left[ 243 ~\right] ~\right\}. $$

Simplifying,

$$f(x_3) = \left(\frac{1}{27}\right) \times ~\left[ -272 + 20\sqrt{10} ~\right] ~< ~0.$$

In summary, for $x$ restricted to negative values, $f(x)$ is maximized at $x = x_3$, with $f(x_3) < 0$. Therefore, there is no value of $x < 0$ such that $f(x) \geq 0.$ Therefore, the function $f(x)$ does not cross the $x$-axis in the interval $-\infty < x < 0.$

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Just another way - to check for negative roots, it is enough to show $x^3+x^2+9=3x$ is not possible for any positive $x$. Using AM-GM, $x^3+(x^2+4)+5 \geqslant x^3+2\sqrt{4x^2}+5=x^3+(4x)+5>3x$, so this isn't possible.

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  • $\begingroup$ how did the question transformed to this "it is enough to show $x^3+x^2+9=3x$ is not possible for any positive x" , cant grasp the logic here of why we are looking for positive x $\endgroup$
    – Fin27
    Nov 5, 2021 at 12:42
  • $\begingroup$ @Fin27 Just consider $f(-x)$, then we are looking for positive values for $x$. You have already considered this same transformation when looking for sign changes on negative axis in your problem. $\endgroup$
    – Macavity
    Nov 5, 2021 at 14:03
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Actually, in working through my Calculus oriented answer, I accidentally thought of a somewhat convoluted precalculus approach to solving the problem.

$f(x)= x^3-x^2-3x-9.$
How many negative real roots does the equation $f(x) = 0$ have ?

$$\text{Let} ~~g(x) = x^3 - x^2 - 3x + 3 = (x - 1) \times (x^2 - 3).\tag1$$


Part of the inspiration for this method, was to construct $g(x)$ such that $f(x) - g(x)$ equals a constant and where $g(x)$ is easy to factor. I noted that the two leftmost coefficients of $f(x)$ followed the pattern $[(+1), (-1)],$ so I constructed $g(x)$ so that the next two coefficients of $g(x)$ would follow the pattern $(-3) \times [(+1), (-1)].$

The other part of the inspiration for this method, was my observation in my other posted answer that the absolute value of the rightmost coefficient of $f(x)$, namely $|-9| = 9$ seemed inordinately large.


Then, with $g(x)$ specified by (1) above, you have that $f(x) = g(x) - 12.$
Therefore, it is sufficient to show that on the interval $-\infty < x < 0$,
the maximum value achieved by $g(x)$ is less than $12$.

Given how $g(x)$ is factored in (1) above, the only values of $x < 0$ for which $g(x)$ will be $ > 0$ will be $-\sqrt{3} < x < 0.$

However, for $-\sqrt{3} < x < 0$, a routine examination of
$|g(x)| = |x - 1| \times |x^2 - 3|$ shows that

  • $|x - 1| = 1 + (-x) < 1 + (\sqrt{3}) ~< ~3 ~< ~4.$
  • $|x^2 - 3| = 3 - x^2 \leq 3.$

Therefore, the product of the two factors above must be strictly less than $(4 \times 3)$. Therefore, the maximum value for $g(x)$ on the interval $-\sqrt{3} < x < 0$ must be less than $(12)$. Since this is the only interval for which $x < 0$ and $g(x) > 0$, the maximum value of $g(x)$ on the interval $-\infty < x < 0$ must be less than $(12)$.

Therefore, since $f(x) = g(x) - 12$, you must have that the maximum value of $f(x)$ on the interval $-\infty < x < 0$ must be less than $0$.

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$x^3 - x^2 - 3x - 9=0$

$3x = x^3 - x^2 - 9$

When $x<0$ every term on the right hand side is less than 0. If we are gong to find equality, $3x$ must be very negative.

When $-3< x < 0, |3x| > |-9|$

When $x< -3, |3x| < |x^2|$ and $|3x| < |x^3|$

There is no way that $3x$ can be sufficiently negative.

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  • $\begingroup$ why are we making cases at x=-3 ? $\endgroup$
    – Fin27
    Nov 5, 2021 at 12:43

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