How to solve what should be a simple variant of a quadratic

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In summary, the conversation revolved around a series of equations that could be solved directly, but the resulting equation had a term (Cx^(-1)) that was causing difficulty. After asking for guidance, the person realized that by multiplying the equation by x, it could be turned into a cubic equation that was more easily solved. It was also noted that while x=0 might work as a solution for the cubic equation, it could not be a solution for the original equation.
  • #1
Justin R
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Hi everyone,

For some reason I can't put this together in my head. I have a series of equations that can be solved directly, but my resulting equation is of the form:

Ax^2 + Bx + Cx^(-1) + D = 0

Obviously the C/x term is the one throwing the wrench in the works here. Could anyone point me in the right direction?

Thanks in advance!

-Justin
 
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  • #2
Justin R said:
Hi everyone,

For some reason I can't put this together in my head. I have a series of equations that can be solved directly, but my resulting equation is of the form:

Ax^2 + Bx + Cx^(-1) + D = 0

Obviously the C/x term is the one throwing the wrench in the works here. Could anyone point me in the right direction?

Thanks in advance!

-Justin

Nevermind... It's very easy. Turn it into a cubic equation by multiplying through by X, resulting in the readily solvable:

Ax^3 + Bx^2 + Dx + C = 0
 
  • #3
And be sure to note that while x= 0 might satisfy the third degree equation, it cannot be a solution to the first equation.
 

FAQ: How to solve what should be a simple variant of a quadratic

How do I identify a quadratic equation?

A quadratic equation is an algebraic equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is also known as a second-degree polynomial equation.

What is the general method for solving a simple variant of a quadratic equation?

The general method for solving a simple variant of a quadratic equation is by using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. This formula is derived from the standard form of a quadratic equation and can be used to find the values of x that satisfy the equation.

How do I determine the number of solutions for a quadratic equation?

The number of solutions for a quadratic equation can be determined by the value of the discriminant, b^2 - 4ac. If the discriminant is positive, there are two distinct solutions; if it is zero, there is one repeated solution; and if it is negative, there are no real solutions.

Can I solve a quadratic equation using factoring?

Yes, quadratic equations can be solved using factoring if the equation can be factored into two binomials. This method involves finding the factors of the constant term (c) and the coefficient of the middle term (b) that add up to the coefficient of the squared term (a). However, not all quadratic equations can be factored, so the quadratic formula may be necessary.

What is the difference between a simple and complex variant of a quadratic equation?

A simple variant of a quadratic equation has one variable (x) and a single squared term (ax^2). A complex variant, also known as a higher-degree polynomial equation, has multiple variables and/or multiple squared terms (ax^2, bx^2, etc.). The methods for solving these two types of equations may differ and complex variants may have multiple solutions.

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