Showing one positive and two neg roots

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The equation x^3 - 12x - 7.2 = 0 has been analyzed to demonstrate it has one positive and two negative roots. By applying Descartes's rule of signs, it is confirmed that there is one positive root and potentially two negative roots. The intermediate value theorem indicates a root exists between f(0) and f(10), reinforcing the presence of one positive root. The function's behavior at infinity and its maxima and minima further illustrate the root distribution. Overall, a graphical sketch shows the function intersects the x-axis at two negative points and one positive point, confirming the root characteristics.
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Show that the equation x^3-12x-7.2=0 has one positive and two negative roots:

I know this can be solved by trial and error, and finding f(0),f(1-4),f(-1 - -4) I have shown that It has two negative roots and one positive however I'm wondering if there is another method on how to show it has two negative and one positive root? I've found the max and min points but I am not sure what to do from there...

thanks
 
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Differentiating the function, you get

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

From this you can observe that the maxima is at -2, and minima is at +2. The function is also continuous. Drawing a rough graph out, you can easily see the type of roots.
 
synkk said:
Show that the equation x^3-12x-7.2=0 has one positive and two negative roots:

I know this can be solved by trial and error, and finding f(0),f(1-4),f(-1 - -4) I have shown that It has two negative roots and one positive however I'm wondering if there is another method on how to show it has two negative and one positive root? I've found the max and min points but I am not sure what to do from there...

thanks
It's a polynomial of odd degree, so we know there is at least one real root.

Descartes's rule of signs could help. It will tell you there is one positive root and perhaps two negative roots.
 
Can you use the intermediate value theorem?

f(0) is negative. f(10) is positive so by the IVT there is a root somewhere in between...
 
Infinitum said:
Differentiating the function, you get

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

From this you can observe that the maxima is at -2, and minima is at +2. The function is also continuous. Drawing a rough graph out, you can easily see the type of roots.

How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.
 
synkk said:
How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.

Well, to draw the graph you would see how the function goes at x = +/- infinity.

\lim_{x\to \infty} f(x) = \infty

and

\lim_{x\to -\infty} f(x) = -\infty

Now, you also know the value of f(0). And you can calculate approximate values(you only need signs) of maxima and minima. Maxima comes out to be positive and minima is negative. Now try completing the graph??
 
synkk said:
How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.

The points on the graph where f(x) intersects x-axis are the roots of the equation.
If you carefully look at the graph, you will see that the graph the negative x-axis at two points and the positive x-axis at one point, hence there are two negative and one positive root.
 
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