Problem 85 - Intermediate Algebra

Question

Suppose that your friend solved the equation \((x+3)^{2}=\) 25 as follows: $$ \begin{aligned} (x+3)^{2} &=25 \\ x^{2}+6 x+9 &=25 \\ x^{2}+6 x-16 &=0 \end{aligned} $$ $$ \begin{aligned} (x+8)(x-2) &=0 & & & \\ x+8 &=0 & \text { or } & & x-2=0 \\ x &=-8 & & \text { or } & & x=2 \end{aligned} $$ Is this a correct approach to the problem? Would you offer any suggestion about an easier approach to the problem?

Step-by-Step Solution

Answer

The approach is correct, but taking square roots is simpler.

Step 1: Identifying Errors in Original Approach

The original approach involves expanding \((x+3)^2\) to \(x^2+6x+9\) accurately, but after setting it equal to 25 and rearranging to form \(x^2 + 6x - 16 = 0\), a quadratic equation is formed and solved using product/sum solving method. The solution is technically correct, but it is unnecessarily complex for this equation by using factoring.

Step 2: Recognizing Simplicity Through Square Root Method

Since the equation \((x+3)^2 = 25\) is already in a perfect square form, we can solve it more simply by taking the square root of both sides. This yields \(x+3 = \pm 5\).

Step 3: Solving for x Using Square Roots

From the equation \(x+3 = \pm 5\), split it into two separate linear equations: \(x+3 = 5\) and \(x+3 = -5\).

Step 4: Finding Solutions of Linear Equations

For \(x+3 = 5\), subtract 3 from both sides to get \(x = 2\). For \(x+3 = -5\), subtract 3 from both sides to get \(x = -8\). Thus giving roots \(x = 2\) and \(x = -8\).

Key Concepts

Factoring Method

The factoring method is one way to solve quadratic equations. It's especially useful when the equation can be easily decomposed into two binomials. By factoring, we aim to express the quadratic equation in the form \[ ax^2 + bx + c = (px + q)(rx + s) = 0 \] The logic behind this approach is that if a product of two numbers is zero, at least one of the numbers must be zero. Thus, we can set each factor equal to zero and solve for the variable.

• Example: Consider the equation \(x^2 + 6x - 16 = 0\).
• We break it down into \((x+8)(x-2) = 0\).
• Next, we solve for \(x\) in each case: \(x+8=0\) or \(x-2=0\).
• Solving these linear equations gives \(x = -8\) and \(x = 2\).

However, even though factoring is a powerful tool, it is not always the most efficient route. There are simpler methods, especially if the quadratic is a perfect square.

Square Root Method

The square root method simplifies solving equations where a perfect square is set equal to a number. It can save time and reduce complexity. This method is ideal if the quadratic is already a quadratic square.To apply the square root method, follow these steps:

• Start: Begin with the equation \((x+3)^2 = 25\).
• Simplify: Take the square root of both sides, which results in \(x+3 = \pm 5\).
• The square root expression \(\pm\) indicates there are two solutions: one positive and one negative.
• Resolve: Break it into two separate equations: \(x+3 = 5\) and \(x+3 = -5\).
• Find x: Solving for \(x\) gives \(x=2\) and \(x=-8\).

In this exercise, the square root method provides a more straightforward solution, avoiding unnecessary complexity.

Solving Linear Equations

Solving linear equations is often the final step after applying methods like factoring or using the square root. Once an equation is reduced to a linear form, solutions become direct.A linear equation in one variable takes the form \(ax + b = 0\). Solving it usually involves simple arithmetic steps:

• Identify: Consider the expression \(x+3 = 5\).
• Isolate x: Subtract 3 from both sides to yield \(x = 2\).
• For \(x+3 = -5\), similarly subtract 3 from each side to find \(x = -8\).

The linear equations give the solutions quickly and efficiently, especially when derived from simplified forms like \(x+3 = \pm 5\). By mastering the fundamental skills of solving linear equations, more complex problems become manageable, setting the stage for solving higher-order problems.