Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation using any convenient method. \((x-3)^{2}=36\)

Step-by-Step Solution

Answer

The solutions are \(x = 9\) and \(x = -3\).

Step 1: Rewrite the Equation

Let's start by rewriting the equation to more easily spot its solutions. The equation, \((x-3)^{2}=36\), can be rewritten as \((x-3)^{2}-36=0\). This means that either \((x-3)^{2} -36 = 0\) or \((x-3)^{2} = 36\).

Step 2: Apply the Square Root Property

The equation \((x-3)^{2} = 36\), means \((x-3)\) can equal either \(+6\) or \(-6\), because a number and its negative both have the same square. Apply these results to obtain two more equations: \(x-3 = 6\) and \(x-3 = -6\).

Step 3: Solve for X

By solving each equation from step 2 separately, two solutions for \(x\) are obtained. For \(x-3 = 6\), adding 3 to both sides gives \(x = 9\). For \(x -3 = -6\), adding 3 to both sides gives \(x = -3\).

Step 4: Confirm the Solutions

As an optional last step, let's confirm these solutions. The original equation to verify is \((x-3)^{2} = 36\). For \(x = 9\), substituting into the problem gives \((9-3)^2 \), which simplifies to \(6^2 = 36\). For \(x = -3\), substituting into the problem gives \((-3-3)^2 \), which simplifies to \((-6)^2 = 36\). Both results confirm that \(x = 9\) and \(x = -3\) are indeed the correct solutions.

Key Concepts

Square Root Method

The square root method is a straightforward way to solve certain quadratic equations like \((x-3)^2 = 36\). The concept relies on understanding that squaring a positive or negative number results in the same square. Here, we start by taking the square root of both sides to isolate the expression.

• For non-variable expressions, this means separating the squared term, giving us two possible equalities.
• Since both positive and negative numbers, when squared, give the same result, we consider both possibilities.

Applying this to our equation, \((x-3)\) could be either equal to \(-6\) or \(6\). Only after figuring out these possibilities can we solve for \(x\).
The square root method is simple and efficient when dealing with such perfect square trinomials.
This method especially shines when the quadratic can be simplified to the form \((variable-term)^2 = constant\).

Solve Quadratic Equation

Solving a quadratic equation like \((x-3)^2 = 36\) involves finding the value(s) of \(x\) that satisfy the equation.Ul>

Start by simplifying or rewriting the equation if necessary. For the given problem, it's already in a convenient form: \((x-3)^2 = 36\).

The goal is to isolate \(x\) and identify its potential solutions. By applying the square root method, we break down the possibilities into simpler equations, \(x-3=6\) and \(x-3=-6\).

The result: two possible solutions for \(x\) found by straightforward algebraic steps.

Quadratic equations have a maximum of two solutions. Each solution indicates a point where the original expression equals zero, emphasizing why understanding the entire process is crucial.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying algebraic expressions to make them easier to work with. In solving \((x-3)^2=36\), there are key steps of manipulation:

• The equation transforms into \((x-3)^2-36=0\), recognizing it can be solved by isolating the squared term.
• Understanding that algebraic manipulation, like moving terms across the equals sign, maintains the equations' equality if done correctly.
• When isolating a term, always ensure to perform the inverse operation on both sides, such as adding 3 to both sides to solve for \(x\) in \(x-3 = 6\) or \(x-3 = -6\).

The power of algebraic manipulation is turning complex equations into solveable, step-by-step processes. When done correctly, it leads to accurate solutions, as verified by substituting the results back into the original equation to check its validity.