Problem 58 - Precalculus

Question

Solve each equation for \(x\) in terms of the other letters. $$(x-p)^{2}+(x-q)^{2}=p^{2}+q^{2}$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = p+q\).

Step 1: Expand Each Square

First, expand the equation \((x-p)^2 + (x-q)^2 = p^2 + q^2\). This gives \((x-p)^2 = x^2 - 2px + p^2\) and \((x-q)^2 = x^2 - 2qx + q^2\). Substitute these into the equation to get: \[ x^2 - 2px + p^2 + x^2 - 2qx + q^2 = p^2 + q^2. \]

Step 2: Combine Like Terms

Combine the terms from both expanded equations. We have:\[ 2x^2 - 2px - 2qx + p^2 + q^2 = p^2 + q^2. \]Reorganize to focus on like terms:\[ 2x^2 - 2(p+q)x + (p^2 + q^2) = p^2 + q^2. \]

Step 3: Simplify the Equation

Subtract \(p^2 + q^2\) from both sides, which cancels these terms out:\[ 2x^2 - 2(p+q)x = 0. \]

Step 4: Factor the Equation

Factor the equation:\[ 2x(x - (p+q)) = 0. \]Now set each factor equal to zero.

Step 5: Solve for x

Set each factor to zero:1. \(2x = 0\) which simplifies to \(x = 0\).2. \(x-(p+q) = 0\) which gives \(x = p+q\).

Step 6: Provide the General Solution

Write the solutions found in Step 5:The values of \(x\) that solve the equation are \(x = 0\) and \(x = p+q\).

Key Concepts

Factoring Equations

Factoring equations involve the process of breaking down an expression into a product of simpler expressions, or 'factors.' This is a critical step in solving complex algebraic equations, and it allows us to solve for unknown variables more efficiently. For the provided equation, after expanding and simplifying, we reached the equation \( 2x^2 - 2(p+q)x = 0 \).
We can factor out the common factor in this expression, which is \( 2x \), leading to:

• \( 2x(x - (p+q)) = 0 \)

By factoring, we decompose the equation into simpler parts—'factors'—that can be individually zero. Understanding this process empowers you to handle more complex polynomials and solve equations by identifying zero points accurately.

Solving for Variables

Once we have factored the equation, the next step is solving for the variable \(x\). Solving for variables is a fundamental concept in algebra that involves finding the values that satisfy the equation. Here, after factoring the equation \( 2x(x - (p+q)) = 0 \), we set each factor equal to zero.
The solutions are:

• For \( 2x = 0 \), simplify to get \( x = 0 \).
• For \( x - (p+q) = 0 \), solve to find \( x = p+q \).

This process demonstrates how factoring simplifies the task of solving for variables, giving us clear potential solutions. It's essential to evaluate each factor separately to ensure all possible solutions are considered. This is particularly important in quadratic and higher-degree equations where multiple solutions may exist.

Quadratic Equations

Quadratic equations are algebraic equations of the second degree, meaning they involve variables raised to the power of two. The general form is \( ax^2 + bx + c = 0 \). In our exercise, during the expansion and combination steps, we deal with terms that fit this structure, especially before the final simplification and factoring. What makes quadratic equations interesting is that they can often produce two distinct solutions, reflecting the parabola's interaction points with the x-axis when graphed.
Quadratics are solved using various methods:

• Factoring: Break down the equation into a product of binomials, like we did above.
• Quadratic Formula: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for more complex equations.
• Completing the Square: Reformat the equation into a perfect square trinomial.

These techniques allow you to determine the roots or solutions for the variable, crucial for understanding the behavior of quadratic functions. In our case, recognizing the quadratic form early aids in selecting the proper method for achieving the solution efficiently.