Problem 57 - Precalculus

Question

Solve each equation for \(x\) in terms of the other letters. $$(a x+b)^{2}-(b x+a)^{2}=0, \text { where } a \neq \pm b$$

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = -1\).

Step 1: Expand Both Sides of the Equation

We start by expanding both sides of the equation, \((ax+b)^2 - (bx+a)^2 = 0\). Let’s use the formula \((p+q)^2 = p^2 + 2pq + q^2\) to expand. - The term \((ax+b)^2\) becomes \(a^2x^2 + 2abx + b^2\).- The term \((bx+a)^2\) becomes \(b^2x^2 + 2abx + a^2\).Substituting back into the equation, we have:\[a^2x^2 + 2abx + b^2 - (b^2x^2 + 2abx + a^2) = 0\].

Step 2: Simplify the Expanded Equation

Now, simplify the equation by combining like terms. Notice that the \(2abx\) terms will cancel each other out.The equation simplifies to:\[a^2x^2 + b^2 - b^2x^2 - a^2 = 0\].This further simplifies to: \[(a^2 - b^2)x^2 + (b^2 - a^2) = 0\].

Step 3: Factor Out Common Terms

Factor out common terms from the simplified equation. Notice that \((a^2 - b^2)\) appears in both terms:\[(a^2 - b^2)(x^2 - 1) = 0\].

Step 4: Solve for x

Now, set each factor equal to zero to solve for \(x\). Since \((a^2 - b^2) eq 0\) because \(a eq \pm b\), we only solve:\[x^2 - 1 = 0\].This gives \(x^2 = 1\) so that \(x = \pm 1\).

Step 5: Conclude the Solution

We found that \(x = \pm 1\) satisfies the equation. Thus, for the given equation, the solutions for \(x\) are \(x = 1\) or \(x = -1\).

Key Concepts

Equation Solving

Equation solving involves finding the value of variables that make the equation true. In our exercise, the goal is to find the values of \(x\) that satisfy the equation \((ax+b)^2 - (bx+a)^2 = 0\). The process involves several important steps:

• First, expand the equation using algebraic manipulations. This involves utilizing formulas for squares of sums, like \((p+q)^2 = p^2 + 2pq + q^2\).
• Next, once expanded, combine like terms and simplify the equation. Look for patterns or symmetrical properties that might help.
• The final step is solving for \(x\), often achieved by factoring or isolating \(x\). In this example, setting each factor from the equation to zero helps identify the values of \(x\).

Equation solving is not just about reaching an answer. Understanding the underlying process enables you to tackle similar problems with confidence. By dissecting complex equations into manageable steps, solving becomes both simpler and more intuitive.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can contain numbers, variables, and operators (such as \(+, -, \times, \div\)). In the context of our exercise, the expressions

• \((ax+b)^2\) and \((bx+a)^2\)

are the building blocks of the equation.
Understanding algebraic expressions involves recognizing their components:

• **Variables:** Symbols like \(x\) in the expression help us generalize and solve equations.
• **Coefficients:** Constant multipliers of the variables, such as \(a\) and \(b\) in this exercise.
• **Constants:** Standalone numbers or symbols, like the \(b^2\) and \(a^2\) in the expanded form.

By breaking down each expression into these parts, we can effectively expand and simplify equations. Algebraic expressions form a foundation upon which we build up to solving the entire problem. This application helps develop a more profound algorithmic thinking necessary for all higher-level math and engineering problems.

Factoring Techniques

Factoring techniques in algebra are critical for simplifying equations and solving them quickly. Factoring involves writing the expression, like a quadratic, as a product of its simpler components. This exercise particularly highlights factoring in its final steps:

• Identify common terms: In the expression \((a^2 - b^2)(x^2 - 1)\), factoring helps collapse the equation to something solvable.
• Recognize special products: The difference of squares \((x^2 - 1)\) factored into \((x-1)(x+1)\) is a classic example.
• Focus on what is non-zero: Since \(a^2 - b^2 eq 0\), knowing this lets us focus solely on solving \(x^2 - 1 = 0\).

Factoring allows us to transform complex equations into simpler ones. This makes finding solutions more manageable, like isolating variables or factoring out expressions. Practicing different factoring techniques builds your toolset, preparing you for a variety of algebraic challenges.