Problem 61 - Algebra 2

Question

Solve for \(x\) in terms of \(a\). $$ 3 x^{2}+a x^{2}=9 x+9 a $$

Step-by-Step Solution

Answer

x = [9 ± sqrt(81 + 36a^2 + 108a)] / (6 + 2a)

Step 1: Combine like terms

Combine the terms containing x^2 on the left side of the equation to simplify the expression. This results in (3 + a)x^2 = 9x + 9a.

Step 2: Rearrange the equation

Move all terms containing x to one side and constant terms to the other side to set the equation to zero. Subtract 9x and 9a from both sides to get (3 + a)x^2 - 9x - 9a = 0.

Step 3: Solve for x

Now we have a quadratic equation in the form Ax^2 + Bx + C = 0, where A = 3 + a, B = -9, and C = -9a. Since we need to solve for x in terms of a, factor the quadratic equation, or use the quadratic formula x = [-B ± sqrt(B^2 - 4AC)] / (2A).

Key Concepts

Quadratic Formula

One of the most reliable methods to solve quadratic equations is by using the quadratic formula, which provides a solution for any quadratic equation of the form \(Ax^2 + Bx + C = 0\). This formula is expressed as \(x = \frac{{-B \pm \sqrt{{B^2 - 4AC}}}}{{2A}}\). As per the given problem, after combining like terms and rearranging the equation, we recognize that \(A = 3 + a\), \(B = -9\), and \(C = -9a\). By substituting these values into the quadratic formula, we can find the value of \(x\) in terms of \(a\).

The beauty of the quadratic formula is in its universality. It can be used for all types of quadratic equations, even when factoring is tricky or impossible. It's particularly useful when dealing with coefficients that are not whole numbers or in equations where the middle term is odd. Students should remember to evaluate the discriminant, \(B^2 - 4AC\), first to check the nature of the roots (real and distinct, real and equal, or complex).

Factoring Quadratics

Factoring quadratics is another technique used when the quadratic equation can be written as a product of two binomials. In the context of our given exercise, after organizing the equation to \( (3 + a)x^2 - 9x - 9a = 0 \), we would look for two numbers that multiply to give \(AC = -9(3 + a)\) and add up to \(B\), which is -9. Factoring can be simpler and quicker than the quadratic formula, but it depends on the equation being 'factorable'.

When teaching this concept, emphasis should be placed on the 'trial and error' approach for finding the correct factors or using techniques like the 'ac method' for more complex equations. This process can sharpen a student's number sense and algebraic understanding, but they should also be acquainted with the limitations of factoring and know when to switch to the quadratic formula.

Combining Like Terms

Combining like terms is a fundamental step in simplifying algebraic expressions. It involves adding or subtracting terms with the same variable and exponent. In our problem, the like terms \(3x^2\) and \(ax^2\) are combined to form \( (3 + a)x^2\). This step is critical because it simplifies the equation and sets the stage for either factoring or applying the quadratic formula.

Understanding how to combine like terms helps students not only in solving quadratic equations but in all areas of algebra. It is a skill that assures cleaner, more manageable equations. As a piece of advice, always look for and combine like terms as one of the first steps when simplifying!

Algebraic Expressions

Algebraic expressions are the backbone of algebra. They consist of numbers, variables, and operators that are combined without an equality sign. In solving our quadratic equation, we manipulate algebraic expressions throughout the process. For example, rearranging terms to one side of an equation to set it to zero is a type of manipulation of these expressions.

Students should learn that algebraic expressions can take many forms, and understanding how to work with them underpins success in algebra. It's also essential to recognize various forms of algebraic expressions, such as binomials and trinomials, which appear commonly in quadratic equations. Mastery of algebraic expressions paves the way for solving more complex problems and is a fundamental skill for success in higher mathematics.