Solving a Quadratic Equation with Unknown X Coefficient: Homework Help

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Homework Help Overview

The discussion revolves around solving the equation x(2x^2 - 5) = -1, which initially appears to be a quadratic equation but is identified as a cubic equation upon expansion. Participants express confusion regarding the presence of the x coefficient outside the parentheses and the nature of the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to expand the equation and collect terms, with some questioning the classification of the equation as quadratic. There are inquiries about how to factor a cubic equation and the implications of the problem being presented in a quadratic context.

Discussion Status

The discussion is ongoing, with participants exploring the nature of the equation and the methods for solving it. Some have provided guidance on expanding the equation and recognizing its cubic form, while others express uncertainty about factoring and solving cubic equations.

Contextual Notes

There is a noted discrepancy regarding the classification of the equation, as it is presented in a unit focused on quadratic equations. Participants also mention that the problem is sourced from a textbook, raising questions about potential errors in the material.

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Homework Statement





Homework Equations



x(2x^2 - 5) = -1

The Attempt at a Solution



This question has been flung at me without any pre-examples how to solve it. I have only dealt with equations like 2x^2 + 5x - 12 = 0. I know that the equation above should be set to zero, but how do I deal with the x coefficient on the paranthesis?

I don't know how to solve this -- it's very confusing and if someone can help me I would be grateful!
 
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I know I need to factorize which is my problem, I am not sure how to when there is an x outside the paramnthesis?
 
Well, the first thing is that this is not a quadratic equation. It is a cubic. Second, when solving polynomial equations, I would first remove all parentheses (by using the distributive law or by multiplying out) and then collect all terms to the left side. Only then would I try to factor, if possible.
 
Do you think there has been a mistake then because the unit I am on is [specifically] quadratic equations? No wonder I wasn't sure about it... If I multiply it out I get

2x^3 - 5x = -1

and then

2x^3 - 5x + 1 = 0

right?
 
So how do I factor that? I've never factored a cubic equation...
 
Yes, the work is correct. Is this problem from a book? There are three solutions, none of which are rational, so I wonder if there was a typo somewhere.
 
It does actually say, make correct to the first decimal place...?
 
and yes, problem from a book... odd to find a cubic equation in the quadratic section...
 
will you please show me how to solve this?
 
  • #10
This is not solvable by factoring. All three roots are irrational, so that means you cannot use the Rational Roots Theorem either. You could just graph it on a graphing calculator and have it find the solutions for you.

There exists a cubic formula, but it is overly complicated to use, and it is not taught AFAIK in courses in elementary/intermediate algebra or precalculus.
 

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