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

In summary, the conversation was about solving a quadratic equation involving fractions and the use of the quadratic formula. The correct coefficients for the equation were discussed and the importance of being careful with variable names in the quadratic formula was emphasized.
  • #1
Cycloned
4
0
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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  • #2
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?
 
  • #3
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?
 
  • #4
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!
 
  • #5
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.
 

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

1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It can be solved using the quadratic formula or by factoring.

2. How do I solve the given quadratic equation?

To solve the equation 2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0, first rearrange the terms to get 2x^2 - (3a + 3b)x + (a^2 + b^2 + 2ab) = 0. Then, use the quadratic formula x = (-b ± √(b^2 - 4ac)) / 2a to find the values of x that satisfy the equation.

3. What is the quadratic formula?

The quadratic formula is a formula used to solve quadratic equations. It states that the solutions to the equation ax^2 + bx + c = 0 are given by x = (-b ± √(b^2 - 4ac)) / 2a.

4. Can the given quadratic equation have more than two solutions?

No, a quadratic equation can have at most two solutions. This is because a quadratic equation is a second-degree polynomial, meaning it has a maximum of two roots.

5. Can I use other methods to solve the quadratic equation?

Yes, in addition to the quadratic formula, you can also solve a quadratic equation by factoring. This involves finding two numbers that, when multiplied, equal the constant term (c) and when added, equal the coefficient of the middle term (b).

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