Problem 53 - Elementary Algebra

Question

\((x+10)^{2}=49\)

Step-by-Step Solution

Answer

The solutions to \( (x+10)^{2}=49 \) are \(x = -3\) and \(x = -17\).

Step 1: Understand the equation

Start by recalling that \(a^{2} = b\) has two solutions: \(a = \pm\sqrt{b}\). In this case, \((x+10)^{2} = 49\), look at it as \(a^{2} = b\) where \(a = x + 10\) and \(b = 49\).

Step 2: Take the square root on both sides

Take the square root of both sides of the equation to obtain \(x+10 = \pm\sqrt{49}\). This simplifies to \(x + 10 = 7\) or \(x + 10 = -7\).

Step 3: Solve each equation

Solve each of the resulting linear equations separately. For \(x + 10 = 7\), subtract 10 from both sides: \ x = 7 - 10 \Rightarrow x = -3 \. For \(x + 10 = -7\), subtract 10 from both sides: \ x = -7 - 10 \Rightarrow x = -17 \.

Step 4: Write the solutions

Combine the solutions from each equation: \(x = -3\) and \(x = -17\).

Key Concepts

Square Roots

Understanding square roots is essential for solving quadratic equations. A square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 49 is 7, because \(7 \times 7 = 49\). Importantly, every positive number has two square roots: a positive and a negative one. Thus, the square roots of 49 are both 7 and -7.

When solving equations like \((x + 10)^2 = 49\), taking the square root of both sides will give you two solutions:

• Positive root: \(x + 10 = 7\)
• Negative root: \(x + 10 = -7\)

This concept is pivotal to tackling quadratic equations since it helps break them down into simpler linear equations.

Linear Equations

Linear equations are equations involving variables with no exponents greater than 1. They form straight lines when graphed and include terms involving the first power of the variable. For example, the equation \(x + 10 = 7\) is a linear equation.

After taking the square roots in our problem, we end up with two linear equations:

• \(x + 10 = 7\)
• \(x + 10 = -7\)

To solve each linear equation, isolate the variable on one side. For example, \(x + 10 = 7\) becomes \(x = 7 - 10\), which simplifies to \(x = -3\). Similarly, \(x + 10 = -7\) becomes \(x = -7 - 10\), which simplifies to \(x = -17\).

By solving these linear equations, you'll find the values of x that satisfy the original quadratic equation.

Algebraic Solutions

Algebraic solutions involve solving equations through various algebraic methods. Quadratic equations like \((x + 10)^2 = 49\) can be broken down using square roots, and then solved using simple algebraic manipulations.

Here's how we approached our problem algebraically:

• Step 1: Recognize that \((x + 10)^2 = 49\) is a quadratic equation.
• Step 2: Take the square root of both sides to get \(x + 10 = \pm\sqrt{49}\).
• Step 3: Simplify the square root of 49 to get \(x + 10 = 7\) or \(x + 10 = -7\).
• Step 4: Solve the resulting linear equations: \(x = -3\) and \(x = -17\).

Combining concepts like square roots and linear equations allows us to solve more complex algebraic problems step by step. Each step builds on fundamental principles that pave the way for mastering algebra.