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

Kulla_9289

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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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Is it y=2/x^2 - x or y=2/(x^2 - x) =\(\displaystyle \dfrac{2}{x^2-x}?\)

It seems that you do have it correct!
 
I do not see it either. So you (not me) need to investigate. Solve 4x^3 - 10x^2 + 2=0 and see if there is a connection between those zeros and the function/graph you are given.

Also, as the posting guideline ask, can you please tell us what topics you have recently learned?
 
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.
I would presume they expect you to find something you can draw on the graph such that its intersection(s) would be the solution(s) to the cubic equation.

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?

I can't be sure if that is what they have in mind, but it does work.
 
As the posting guideline ask, can you please tell us what topics you have recently learned?
 
I would presume they expect you to find something you can draw on the graph such that its intersection(s) would be the solution(s) to the cubic equation.

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?

I can't be sure if that is what they have in mind, but it does work.
How would I do that?
 
How would I do that?

You need to start with...
[math]\frac{2}{x^2} - x = ax + b[/math]and manipulate it so that it looks like...
[math]Ax^3 + Bx^2 + D = 0[/math]
A,B, and D will be in terms of the variables a & b (or just a constant)
Then you'll need to find values for "a" and "b" such that A=4, B=-10, and D=2

Start by eliminating the x^2 in the denominator. To accomplish this you must do something to both sides of the equation.
 
2/x^2−x=ax+b
To get a cubic, just multiply by x^2 to get 2 -x^3 = ax^3 + bx^2
 
Confused
 

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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
 
How is [imath]2/x^2*x^2-x=2-x^3[/imath]? Isn't it [imath]2-x[/imath]?
You need to multiply both entire sides by [imath]x^2[/imath], not just one term. So you want [imath](2/x^2-x)*x^2[/imath], that is, [math]\left(\frac{2}{x^2}-x\right)\cdot x^2[/math].

What do you get for that?
 
@Dr.Peterson So that's the rule. I haven't an idea as to why I have been taught [imath]3x=1/2*2[/imath] for [imath]3x=1/2[/imath] instead of [imath]3x=(1/2)*2[/imath]. These inconsistencies that have been impressed upon our minds are inextricable. Practises upon practises have been committed just to impress this upon our minds.

Makes sense now.
 
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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
Please elaborate. I do not understand why.
 
@Dr.Peterson So that's the rule. I haven't an idea as to why I have been taught [imath]3x=1/2*2[/imath] for [imath]3x=1/2[/imath] instead of [imath]3x=(1/2)*2[/imath]. These inconsistencies that have been impressed upon our minds are inextricable. Practises upon practises have been committed just to impress this upon our minds.

Makes sense now.
I'm not quite sure what you are saying here. To solve [imath]3x=1/2[/imath] you would not multiply (only) the right side by 2; you would multiply both sides by [imath]\frac{1}{3}[/imath] to isolate the variable by eliminating its coefficient.

And the two things you wrote, "[imath]3x=1/2*2[/imath] ... instead of [imath]3x=(1/2)*2[/imath]" are identical.

The main point I made that you said makes sense is that you multiply both sides by the same thing, and in particular all of each side. The key idea is that if two quantities are equal, then the results of multiplying them both by the same thing are still equal -- you get an equivalent equation.

Please elaborate. I do not understand why.
How are you doing on the problem as a whole? Which part of what Steven G (or others) said do you not understand?

Also, I want to ask about your own personal context: How much of algebra should we expect you to know? 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.
 
@Dr.Peterson How did [imath]ax^3[/imath] transpose into [imath](a+1)x^3[/imath]? And why? And what is equating coefficient? And how did 0 and -2 come from in (a+1)x^3 + bx^2 -2 =0?
 
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