Solve Quadratic Equation: 2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0

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Homework Help Overview

The discussion revolves around solving a quadratic equation derived from a problem involving fractions. The original poster expresses difficulty in reaching a solution and presents the equation 2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation into a standard quadratic form and question the correctness of coefficients. Some participants inquire about the quadratic formula and its application, while others highlight potential misinterpretations of the original problem setup.

Discussion Status

The discussion is ongoing, with participants providing insights into the structure of the quadratic equation and questioning the original poster's interpretation. There is an acknowledgment of the need for clarity regarding variable definitions and constraints related to the values of x.

Contextual Notes

Participants note that the equation involves division by expressions that could lead to undefined values, specifically indicating that x cannot equal a or b. There is also mention of potential errors in the coefficients of the quadratic equation presented by the original poster.

Cycloned
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Need help solving some. I'll put up one for now.

a/b-x + b/a-x = 2

I'm just looking for how to solve it, because I always reach

2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0

and get stuck. Please show me your steps.

Thanks!
 
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2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0

That = 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0

Do you know the quadratic equation?
 
I know quadratic equations, but I'm getting stuck here, because I do not know how to solve for x.

I did establish the fact that 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0
is correct because it follows ax^2 + bx + c.

Could you please help me out?
 
Cycloned said:
Need help solving some. I'll put up one for now.

a/b-x + b/a-x = 2
What this means is probably different from what you intended. When you write fractions in a single line of text, use parentheses if the numerator or denominator have more than one term.

What you wrote would be interpreted as this:
[tex]\frac{a}{b - x} + \frac{b}{a - x} = 2[/tex]
Cycloned said:
I'm just looking for how to solve it, because I always reach

2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0
I get something different - different coefficients for the two terms in x, and different signs for the a^2 and b^2 terms. Check your work.
Cycloned said:
and get stuck. Please show me your steps.

Thanks!
 
l'Hôpital said:
That = 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0

Do you know the quadratic equation?

2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0 IS a quadratic equation. The question should be: Do you know the quadratic formula?

The quadratic formula gives solutions to the equation ax^2 + bx + c = 0. Since both the quadratic formula and the equation above use a and b, be careful in what you call a, b, and c in the quadratic formula.

As already noted, some of the coefficients in the OP's equation are incorrect.

Also, since the equation in the original post involved division by a - x and b - x, it should be stated explicitly that x can't be a, and x can't be b.
 

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