Use the graph of y=2/x^2 - x below to solve the equation 4x^3-10x^2+2=0.

Dr.Peterson said:
We generally guess based on the level of a problem, but this one seems far beyond someone who has trouble solving a linear equation with a fraction in it, so we may be talking over your head. You answered just a little bit of this in #7.
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It’s because @Steven G didn‘t explain properly; you should always explain why you would do this (and why take this particular approach). I know what I am doing so worry not.
 
Kulla_9289 said:
It’s because @Steven G didn‘t explain properly; you should always explain why you would do this (and why take this particular approach). I know what I am doing so worry not.
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[math]\frac{2}{x^2} - x = ax + b\\ \text{Multiply both sides by } x^2:\\ \red{x^2}\left(\frac{2}{x^2} - x \right)= \red{x^2}(ax + b)[/math]
Can you continue?
 
Kulla_9289 said:
It’s because @Steven G didn‘t explain properly; you should always explain why you would do this (and why take this particular approach). I know what I am doing so worry not.
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This is not always true. I knew that my hint was somewhat vague. I just want you to think and fill in the missing steps. Getting a student to think mathematically on their own is a far greater value to the student then just giving the student the solution. This is the approach that almost all helpers use on this forum.
 
BigBeachBanana said:
[math]\frac{2}{x^2} - x = ax + b\\ \text{Multiply both sides by } x^2:\\ \red{x^2}\left(\frac{2}{x^2} - x \right)= \red{x^2}(ax + b)[/math]
Can you continue?
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[imath]2-x^3[/imath][imath]=ax^3 +bx^2[/imath]. What are we trying to accomplish?
 
Steven G said:
2 -x^3 = ax^3 + bx^2 => (a+1)x^3 + bx^2 -2 =0 and Ax^3 + Bx^2 + D =0
Equate coefficients for (a+1)x^3 + bx^2 -2 and Ax^3 + Bx^2 + D

A=a+1, B =b and D=-2
You are given A, B and D
Continue
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@Steven G you did the following: to achieve [imath]Ax^3+Bx^2 +D[/imath], you wanted to eliminate 2 by subtracting 2 both sides. That‘s how we got [math]ax^3 + bx^2 -2[/math]. But I still don’t understand why you had factorised.
 
Kulla_9289 said:
Use the graph of y=2/x^2 - x below to solve the equation 4x^3-10x^2+2=0. I literally have no idea as to what to do.
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Here is most of the work.
Let 2/x^2 - x = ax+b
By multiplying both sides by x^2 we get 2-x^2 = ax^3 + bx^2 or -ax^3 - (b+1)x^2 + 2 =0
Set -ax^3 - (b+1)x^2+2 = 4x^3-10x^2+2
Then a=-4, b+1=10,
So a=-4 and b=9
 
Steven G said:
Here is most of the work.
Let 2/x^2 - x = ax+b
By multiplying both sides by x^2 we get 2-x^2 = ax^3 + bx^2 or -ax^3 - (b+1)x^2 + 2 =0
Set -ax^3 - (b+1)x^2+2 = 4x^3-10x^2+2
Then a=-4, b+1=10,
So a=-4 and b=9
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Check that. Others had the equation right.

In fact, you did, in #14:
Steven G said:
2 -x^3 = ax^3 + bx^2 => (a+1)x^3 + bx^2 -2 =0 and Ax^3 + Bx^2 + D =0
Equate coefficients for (a+1)x^3 + bx^2 -2 and Ax^3 + Bx^2 + D

A=a+1, B =b and D=-2
You are given A, B and D
Continue
Click to expand...
 
Kulla_9289 said:
So, x^3(a+1) + bx^2 - 2
Click to expand...
Kulla_9289 said:
What happens to the D? Are we drawing a straight line?
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As was stated in #14, you have found that

(a+1)x^3 + bx^2 - 2 = Ax^3 + Bx^2 + D​

That gives us expressions for A, B, and D.

And why did we do that? Because, as I said in #5,
Dr.Peterson said:
So, if you drew a line y = ax + b, what would it take for the equation 2/x^2 - x = ax + b to be equivalent to 4x^3 - 10x^2 + 2 = 0?
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You want the intersections where 2/x^2 - x = ax + b to be the places where 4x^3 - 10x^2 + 2 = 0.

What do A, B, and D have to be?

Actually, it's a little confusing to say it that way. Rather than talk about A, B, and D, which were just a way to say "put it in standard form for a polynomial" let's focus on the original goal.

What you've really done is to show that the equation 2/x^2 - x = ax + b, representing intersections of the given curve and the line we are trying to define, is equivalent to

(a+1)x^3 + bx^2 - 2 = 0​

We want this to be equivalent to the given equation

4x^3 - 10x^2 + 2 = 0​

Note that the constant terms (-2, 2) are different; so change our equation to

-(a+1)x^3 - bx^2 + 2 = 0​

For two polynomials to be equivalent, their coefficients have to be the same; so we need

-(a+1) = 4​
-b = -10​
2 = 2​

Use that to find a and b, and you'll have a line to draw on the graph you were given.
 
@Dr.Peterson how would I solve 3x^2 – 7 = 0 if I am given a graph of y = 3x^2 – x – 2?
So, 3x^2 – x – 2 = ax + b. I would need to form that equation into this form Ax^2 - D = 0. What do I need to afterwards?
 
Kulla_9289 said:
@Dr.Peterson how would I solve 3x^2 – 7 = 0 if I am given a graph of y = 3x^2 – x – 2?
So, 3x^2 – x – 2 = ax + b. I would need to form that equation into this form Ax^2 - D = 0. What do I need to afterwards?
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In my opinion, this does belong in this thread, since it is the same kind of problem, and you just need to apply the same method, so seeing them together helps us.

But you should just do one step at a time! Put this equation into the form Ax^2 + Bx + C (that is, a general quadratic), and equate it to 3x^2 – 7 = 0 as before. (Note that in the previous problem, I would have talked about Ax^3 + Bx^2 + Cx + D, not Ax^3 + Bx^2 + D, and this example illustrates why.)

Show us your results, and if it isn't clear what to do next, we can help you. Problem solving should not be dependent on having a complete plan, or on being told everything to do.
 
So, 0 = ax + b + x + 2 - 3x^2. I am really not sure now. Do I need to factorise? (a + 1 - 3x)x + b + 2?
 
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