Doubt about sum of roots

[hail385]

We are learning quadratic equations and expressions, and this question was given in an assignment.


[-7/8 is the given answer]

This is my answer.



My doubts about this are:

1. This answer should be valid for the points intercepted by any line of the form y=c. However, will the arithmetic mean of the x-coordinates still be -7/8 still be if we consider a the points intercepted by a line that is not parallel to the x-axis.?
2. Is the answer actually -7/8?

To get an answer to 1, I entered the equation in Desmos, and this was shown.​

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There were only two points that actually lied on the x-axis.​

Desmos | Graphing Calculator

www.desmos.com

And the mean of the x-coordinates was not equal to -7/8.​

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Moreover, if we look at the equation [math]2x^4+7x^3+3x-5=0[/math], then the sum of the roots must be [math]-7/2 = -3.5[/math]​

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But after finding the solutions to the equations from Symbolab, they did not add up to -3.5. Since a quartic equation should have 4 roots, I found out what the sum would be in case two of the roots were 0.701 and also other variations, but the sum still did not add up to -3.5. There are no complex roots to this equation either.

Why does this happen?

Also, is it necessary that there should be a line that intercepts the curve [math]y=2x^4+7x^3+3x-5[/math] at four distinct points .Can you please give an example?

Or is there something wrong with my solution?

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[Deleted member 4993]

This answer should be valid for the points intercepted by any line of the form y=c.

Why are you assuming that the intercepting line is parallel to the x-axis?

Your OP states that some collinear points of interception. So instead of y=c you should assume y = mx +c.

 

[BigBeachBanana]

There are no complex roots to this equation either.

Complex roots do exist for this equation. To add to post#2, you graphed the function yourself and saw only two roots, not 4.

 

[blamocur]

Why are you assuming that the intercepting line is parallel to the x-axis?

Your OP states that some collinear points of interception. So instead of y=c you should assume y = mx +c.

In fact there no horizontally collinear points for this graph:

 

[hail385]

Complex roots do exist for this equation. To add to post#2, you graphed the function yourself and saw only two roots, not 4.

I thought Symbolab usually showed complex solutions if any
I think I missed this line. So this means that there are complex roots that symbolab can't find?

Why are you assuming that the intercepting line is parallel to the x-axis?

Your OP states that some collinear points of interception. So instead of y=c you should assume y = mx +c.

OK yes. I think I took the y=0 case because I was taught to use the simplest cases for such problems and use as little variables as possible because it might make solving the problem more complex and got too used to it. I solved it in my notebook before checking on Desmos. Yes the arithmetic mean would be -7/8 for any other line too. Thanks.

And I also managed to find a line that does intercept the curve at 4 points using desmos. And in this case the arithmetic mean is -7/8.
I realized that I wouldn't have had such doubts if I thought more about it and checked the process carefully. Sorry for taking your time and thanks for answering.

 

[Dr.Peterson]

This answer should be valid for the points intercepted by any line of the form y=c. However, will the arithmetic mean of the x-coordinates still be -7/8 still be if we consider a the points intercepted by a line that is not parallel to the x-axis.?

The work shows that the mean of the four roots is always -7/8. And since the work (which is wrong only in claiming that any four collinear points must be x-intercepts) doesn't depend on the linear and constant terms, it also applies to the roots of 2x^4+7x^3+3x−5 = ax+b; that is, to any four collinear points on the graph.

Is the answer actually -7/8?

Yes.

And the mean of the x-coordinates was not equal to -7/8. Moreover, if we look at the equation 2x^4+7x^3+3x-5=0, then the sum of the roots must be −7/2=−3.5

And therefore the mean is that divided by 4, which is -7/8.

If you included the non-real roots, it would work. You only found the mean of two of them.

But after finding the solutions to the equations from Symbolab, they did not add up to -3.5. Since a quartic equation should have 4 roots, I found out what the sum would be in case two of the roots were 0.701 and also other variations, but the sum still did not add up to -3.5. There are no complex roots to this equation either.

Why does this happen?

There are complex roots. Why do you think there are none? And the two real roots are not double roots, because the curve is not tangent to the axis. You can find the complex roots by dividing the polynomial by each of the real roots you found (approximately).

Also, is it necessary that there should be a line that intercepts the curve y=2x^4+7x^3+3x-5 at four distinct points .Can you please give an example?

Not all lines will intersect in four points; your graph shows that! But the problem says, IF there are four collinear points; the work applies in any such case. And the answer holds even if there is no such line.

Although not all quartics have four collinear points, here is one such line (y = 15x - 5) for your polynomial:

 

[hail385]

the curve is not tangent to the axis

I dont think my teacher has taught this part yet but I will keep it in mind.

Although not all quartics have four collinear points,

Thanks for telling me about this too.

I apologise that you had to take the time to write all of that. I realized my mistakes earlier but since I’m a new member, my messages have to be approved before they could be displayed, during which time you answered.

 

[hail385]

You can find the complex roots by dividing the polynomial by each of the real roots you found (approximately).

I didn’t understand this part. Can you please explain how it works?

 

[Dr.Peterson]

I didn’t understand this part [complex roots by dividing]. Can you please explain how it works?

That's what Symbolab explained in its first couple steps you showed in #5. Find one root, such as 0.70117, then divide by (x - 0.70117) to get a cubic; then find a root of that, such as -3.66269, and divide by (x + 3.66269) to get a quadratic. But then, you can find the two complex roots of that quadratic; all they did was try to find a real root, which doesn't exist.