Which arithmetic mean of x-coordinates for collinear points on y=2x^4+7x^3+3x-5?

[utkarshakash]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I think that the four points mentioned must be the roots of the equation.

 

[SammyS]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I think that the four points mentioned must be the roots of the equation.

Graph the function. It only has one real root.

 

[sjb-2812]

Graph the function. It only has one real root.

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)

 

[Curious3141]

Graph the function. It only has one real root.

There are 2 real roots.

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I think that the four points mentioned must be the roots of the equation.

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

 

[SammyS]

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

Is that correct?

 

[utkarshakash]

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

Is that correct?

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

 

[utkarshakash]

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

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?

 

[SammyS]

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.

 

[ehild]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I think that the four points mentioned must be the roots of the equation.

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

 

[utkarshakash]

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

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

 

[utkarshakash]

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

Can you please help me on the other question of mine posted by the title
'Find least value of a'?