Problem 37 - Precalculus : Building Concepts and Connections

Question

Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{x^{2}+1}=\sqrt{17}$$

Step-by-Step Solution

Answer

The real solutions to the given equation are \(x=4\) and \(x=-4\).

Step 1: Square both sides of the equation

To remove square roots, square both sides of the equation, which yields \(x^{2}+1=17\).

Step 2: Solve for x

Next, move 1 to the right side of the equation to isolate \(x^{2}\) on the left, resulting in \(x^{2}=17-1=16\). Now, take the square root on both sides to find the possible values of \(x\). Therefore, \(x =\sqrt{16}\) or \(x =-\sqrt{16}\). This results in the two possible solutions \(x=4\) and \(x=-4\).

Step 3: Check the solutions

Substitute each solution back into the original equation to ensure they are valid, not extraneous solutions. For \(x=4\), \(\sqrt{(4)^{2}+1}=\sqrt{17}\), which is true so \(x=4\) is a valid solution. For \(x=-4\), \(\sqrt{(-4)^{2}+1}=\sqrt{17}\), which is also true so \(x=-4\) is a valid solution.

Key Concepts

Solving Radical Equations

Radical equations are equations that involve roots, often square roots. Solving them requires isolating the radical term and then eliminating it, usually by squaring both sides of the equation. This approach can help reveal the underlying relationship between the variables involved.
To solve the equation \( \sqrt{x^2 + 1} = \sqrt{17} \), we start by squaring both sides. This action removes the square root, which simplifies the equation to \( x^2 + 1 = 17 \).

• Isolate the variable: Subtract 1 from both sides to focus on \( x^2 \), simplifying it to \( x^2 = 16 \).
• Solve for \( x \): Finding the square root on both sides gives us the potential values, \( x = 4 \) or \( x = -4 \).

By following these steps, we discover two possible solutions. However, not every solution we calculate is valid, which brings us to the next crucial step: validating solutions.

Validating Solutions

Once we have potential solutions, it's essential to confirm they truly satisfy the original equation and aren't extraneous solutions. Extraneous solutions arise when we manipulate equations, like squaring, because these operations can introduce invalid results that don't meet the original equation's requirements.
To validate solutions for our given problem, check both \( x = 4 \) and \( x = -4 \) by substituting them back into the initial equation \( \sqrt{x^2 + 1} = \sqrt{17} \):

• For \( x = 4 \), substitute to get \( \sqrt{(4)^2 + 1} = \sqrt{17} \). This simplifies to \( \sqrt{17} = \sqrt{17} \), confirming \( x=4 \) is valid.
• For \( x = -4 \), substitute to get \( \sqrt{(-4)^2 + 1} = \sqrt{17} \). This also simplifies to \( \sqrt{17} = \sqrt{17} \), confirming \( x=-4 \) is valid.

Validating solutions ensures we're only considering real answers that work within the original context of the problem.

Real Solutions of Radical Equations

The concept of real solutions refers to finding solutions that are actual numbers on the real number line, not imaginary or complex numbers. In the context of radical equations, we must be careful because the manipulations used to solve them can sometimes hint at solutions that aren't truly applicable. The equation \( \sqrt{x^2 + 1} = \sqrt{17} \) is inherently about finding values where this equation holds true in the real number realm. Both results, \( x=4 \) and \( x=-4 \), are examples of real solutions because they satisfy the original equation when checking their validity. In solving radical equations, it's crucial to:

• Consider only real roots: Although solving \( x^2 = 16 \) suggests positive and negative roots, not all equations guarantee real solutions.
• Examine domain restrictions: Radical expressions can restrict domains, but in this problem, all values found were valid in the reals.

Understanding these concepts helps learners discern which solutions truly apply, ensuring accuracy in applying mathematical skills to real-world and theoretical problems.