# Thread: cubic observation

1. ## cubic observation

I have been playing around with some geogebra and noticed this and i can't see if it is general or not...

Take any cubic equation with a turning points and find its three points of intersection with a line ( i.e. line cuts through the curve in 3 places). If i sum of the x values of the points of intersection , it comes to a number , say p. If i then move the line to another place by messing with its gradient or intercept ( where it still cuts in 3 places), the sum of the x values is p again!

Does this make any sense?

2. ## Re: cubic observation

So you have a cubic polynomial: $p(x) = (x-a)(x-b)(x-c)$. You are setting it equal to a line $l(x) = mx+k$. This gives, $p(x)=l(x)$, so $(x-a)(x-b)(x-c) = mx+k \Longrightarrow x^3-(a+b+c)x^2+(ab+ac+bc-m)x-(abc+k) = 0$. You are saying that regardless of what you use for $m$ or $k$, so long as the modified polynomial has three real roots, $a_1,b_1,c_1$ for one choice of $m_1,k_1$ and $a_2,b_2,c_2$ for another choice of $m_2,k_2$, then $a_1+b_1+c_1 = a_2+b_2+c_2$?

3. ## Re: cubic observation

Originally Posted by SlipEternal
So you have a cubic polynomial: $p(x) = (x-a)(x-b)(x-c)$. You are setting it equal to a line $l(x) = mx+k$. This gives, $p(x)=l(x)$, so $(x-a)(x-b)(x-c) = mx+k \Longrightarrow x^3-(a+b+c)x^2+(ab+ac+bc-m)x-(abc+k) = 0$. You are saying that regardless of what you use for $m$ or $k$, so long as the modified polynomial has three real roots, $a_1,b_1,c_1$ for one choice of $m_1,k_1$ and $a_2,b_2,c_2$ for another choice of $m_2,k_2$, then $a_1+b_1+c_1 = a_2+b_2+c_2$?
Ah, so because the coefficient of x^2 is not affected by the linear mx+k, the sum of roots remains the same! Nice .

4. ## Re: cubic observation

He was actually asking a question, but he has more or less given the answer.

Your cubic has an equation $p(x) = x^3 + px^2 + qx + r = (x-a)(x-b)(x-c)$ where $a$, $b$, and $c$ are the x-coordinates at which the cubic crosses the x-axis (which is just a line $y=0x+0$. By equating coefficients we find that $p = a + b + c$.

Now, when you draw another line and investigate the intersection points with the general line $y=mx+k$, you get $x^3 + px^2 + qx + r = mx+k$ which means (as above) that $x^3 + px^2 + (q-m)x + (r-k) = 0$. This equation describes the intersection of another cubic with the x-axis. Namely $p_1(x) = x^3 + px^2 + (q-m)x + (r-k) = (x-a_1)(x-b_1)(x-c_1)$ where $a$, $b$, and $c$ are the x-coordinates at which the cubic crosses the x-axis and also the x-coordinates of where your line intersects the original cubic. And by equating coefficients again we get $p = a_1 + b_1 + c_1$. Thus, the conclusion that you drew.