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

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 ?

• 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. Nov 4, 2021 at 7:43
• @user2661923 , you can post the solution, there's no harm in learning different approaches :) Nov 4, 2021 at 8:26
• ...except the two roots other than $\ 3 \$ are complex $\ ( -1 \ \pm \ i·\sqrt2 ) \ \ , \$ not purely imaginary.
– user882145
Dec 9, 2022 at 6:07

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

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

• How were you able to factorise the equation? Nov 4, 2021 at 18:01
• rational root theorem Nov 4, 2021 at 18:17

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 $$-\infty < x < 0.$$

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.

• 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 Nov 5, 2021 at 12:42
• @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. Nov 5, 2021 at 14:03

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 $$0$$.

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

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

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.

• why are we making cases at x=-3 ? Nov 5, 2021 at 12:43