# Homework Help: Find the arithmetic mean

1. Nov 1, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
If four distinct points on the curve $y=2x^4+7x^3+3x-5$ are collinear, then find the arithmetic mean of x-coordinates of the aforesaid points.

2. Relevant equations

3. The attempt at a solution
I think that the four points mentioned must be the roots of the equation.

2. Nov 1, 2012

### SammyS

Staff Emeritus
Graph the function. It only has one real root.

3. Nov 1, 2012

### sjb-2812

Are you sure? I thought for any polymonial, complex roots come in even numbers, so that quartics have 0, 2 or 4 real roots (some may be repeated, mind)

4. Nov 1, 2012

### Curious3141

There are 2 real roots.

5. Nov 1, 2012

### Ray Vickson

I assume you mean that for f(x) = 2x^4+7x^3+3x-5, the four points (x1,f(x1)), (x2,f(x2)),(x3,f(x3)), (x4,f(x4)) lie on a straight line. If you plot f(x) you will see that the places where you can have 4 collinear points on the curve y = f(x) are quite limited (although there are still infinitely many possibilities). I don't see why the roots of f(x) have much to say about this problem.

RGV

Last edited: Nov 1, 2012
6. Nov 1, 2012

### SammyS

Staff Emeritus
I assume that since this thread is posted in the Precalculus Mathematics section, you are not to use calculus.

Is that correct?

7. Nov 1, 2012

### utkarshakash

A liitle bit of it won't do any harm.

8. Nov 1, 2012

### utkarshakash

I plotted the graph and it comes out that there are only 2 real roots. So the points mustn't be the roots. Then what are those 4 points?

9. Nov 1, 2012

### SammyS

Staff Emeritus
Inflection points for $y=2x^4+7x^3+3x-5$ occur where the second derivative is zero.

These occur at (-7/4, -3713/128) and at (0, -5).

The line passing through these points has the equation, y = (439/32)x-5 .

Subtracting (439/32)x-5 from your function gives, $\displaystyle y=2 x^4+7 x^3-\frac{343}{32}x\ .$

Here's the graph of that from WolframAlpha.

Of course two of the zeros are at the inflection points. You can then find the other two for the remaining quadratic.

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• ###### WA_Degree4_Poly.gif
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10. Nov 1, 2012

### ehild

The four points are on a straight line y=ax+b, and are points of the curve at the same time. That means: $ax+b=2x^4+7x^3+3x-5$
The x coordinates of the collinear points are roots of the rearranged equation $2x^4+7x^3+(3-a)x-5-b=0$, and the equation has four real roots.

You need the geometric mean of the x coordinates, which is the sum of roots divided by 4. And you know the relation between roots and coefficients.

ehild

11. Nov 2, 2012

### utkarshakash

You are a genius. Thanks for helping. There was no need of those calculus methods which were extremely complicated.

12. Nov 2, 2012