# Thread: Show that the cubic equation has one positive and two negative roots. How?

1. ## Show that the cubic equation has one positive and two negative roots. How?

Show that the equation x^3-12x-7.2=0 has one positive and two negative roots. Obtain the positive root to 3 significant figures using the Newton-Raphson process.
How do i do the blue part. I am clueless.

Show that the equation x^3-12x-7.2=0 has one positive and two negative roots. Obtain the positive root to 3 significant figures using the Newton-Raphson process.
How do i do the blue part. I am clueless.
Let f(x) = x^3 - 12x - 7.2

Note that f(x) is continuous for all values of x. Apply the Intermediate Value Theorem:

f(0) < 0 and f(4) > 0 => f(a) = 0 for 0 < a < 4.

f(-4) < 0 and f(-3) > 0 => f(b) = 0 for -4 < b < -3.

f(-3) > 0 and f(-1/2) < 0 => f(c) = 0 for -3 < c < -1/2.

Therefore two negative roots x = b and x = c and one positive root x = a.

3. let's look at the derivative of this function:

$f'(x) = 3x^2 - 12$

as you can see the function has to extrema points: a minima at (2,-23.2) and a maxima at x = (-2,8.8).

We can also observe that:
$

\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } {\kern 1pt} \,f(x) = \infty \hfill \\
\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \,f(x) = - \infty \hfill \\
\end{gathered}
$

and also that
$f(0) = - 7.2$

Thus using the aforementioned facts and the fact that the function is continues one can easily conclude that f(x) has one negative root between 0 and -2, another negative root that's smaller than -2, and a positive root larger than 2.

4. Originally Posted by Peritus
let's look at the derivative of this function:

$f'(x) = 3x^2 - 12$

as you can see the function has to extrema points: a minima at (2,-23.2) and a maxima at x = (-2,8.8).

We can also observe that:
$

\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } {\kern 1pt} \,f(x) = \infty \hfill \\
\mathop {\lim }\limits_{x \to - \infty } {\kern 1pt} \,f(x) = - \infty \hfill \\
\end{gathered}
$

and also that
$f(0) = - 7.2$

Thus using the aforementioned facts and the fact that the function is continues one can easily conclude that f(x) has one negative root between 0 and -2, another negative root that's smaller than -2, and a positive root larger than 2.
Less trial and error. I will use that method then.