Problem 22 - College Algebra with Corequisite Support

Question

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. $$ \sqrt{x-1}=x-7 $$

Step-by-Step Solution

Answer

The solution is \(x = 10\).

Step 1: Eliminate the square root

To eliminate the square root, square both sides of the equation. This yields: \[ (\sqrt{x-1})^2 = (x-7)^2 \ x - 1 = x^2 - 14x + 49 \]

Step 2: Rearrange into a standard quadratic equation

Move all terms to one side of the equation so it is in the form \(ax^2 + bx + c = 0\). Start by moving \(x\) and \(1\) to the right side:\[ 0 = x^2 - 14x + 49 - x + 1 \ 0 = x^2 - 15x + 50 \]

Step 3: Factor the quadratic equation

Factor the quadratic equation. The equation \(x^2 - 15x + 50\) factors as:\[(x - 10)(x - 5) = 0\]

Step 4: Solve the factored equation

Set each factor equal to zero and solve for \(x\):1. \(x - 10 = 0\) gives \(x = 10\)2. \(x - 5 = 0\) gives \(x = 5\)

Step 5: Check for extraneous solutions

Substitute both possible solutions back into the original equation \(\sqrt{x-1} = x-7\) to check if they are valid:- For \(x = 10\): \[\sqrt{10-1} = 10-7 \ \sqrt{9} = 3\] This is valid.- For \(x = 5\): \[\sqrt{5-1} = 5-7 \sqrt{4} = -2\]This is not valid because \(\sqrt{4} = 2\), not \(-2\).Thus, \(x = 5\) is an extraneous solution.

Key Concepts

Quadratic Equations

Quadratic equations are a popular topic in algebra that involve polynomials of degree 2. They usually come in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations are called "quadratic" because "quad" refers to squaring, which is central to their structure.
Quadratic equations can be solved using various methods like factoring, completing the square, or using the quadratic formula. However, in this exercise, we focused on solving a radical equation by transforming it into a quadratic equation.

• To solve, we first needed to eliminate the square root by squaring both sides of the equation.
• This conversion gave us a standard quadratic equation to work with.

Once in a quadratic form, we can follow familiar algebraic techniques to find its solutions.

Factoring Polynomials

Factoring polynomials is an efficient method for solving equations, especially quadratics. It involves expressing a polynomial as a product of its factors.
For example, the quadratic \( x^2 - 15x + 50 \) can be broken down into \((x - 10)(x - 5) = 0\).
Factoring is favorable because it simplifies the equation, allowing us to find the solutions easily:

• Set each factor equal to zero and solve for \( x \).
• For \( (x - 10) \), solving gives \( x = 10 \).
• For \( (x - 5) \), solving gives \( x = 5 \).

When polynomial factoring works, it often leads to faster results. However, being adept at recognizing how to factor is a crucial skill that comes with practice.

Extraneous Solutions

Extraneous solutions are solutions that arise from the process of solving the equation but do not satisfy the original equation. They often occur when both sides of an equation are squared, as seen in radical equations.
Checking for these types of solutions is an essential step. To confirm whether a solution is extraneous, substitute each solution back into the original equation:

• For the solution \( x = 10 \), substituting back remains valid.
• For \( x = 5 \), however, it does not satisfy the original equation \( \sqrt{x-1} = x-7 \) as we ended up attempting to equate \( \sqrt{4} = -2 \).

Thus, the solution \( x = 5 \) is extraneous. Identifying extraneous solutions ensures we are considering only valid solutions in our final answer.

Algebraic Equations

Algebraic equations are mathematical statements that use algebra to express relationships between different variables and constants. Solving such equations involves finding the value of the unknown variable that makes the equation true.
This exercise showcased how we can simplify a radical equation into an algebraic one before progressing into solving it step by step. The ultimate aim is to isolate the variable - in this case, \( x \) - and determine its value(s).

• Begin by removing radicals to deal with polynomial expressions.
• Manipulate the equation to gather similar terms on one side to facilitate solving.

Through understanding and applying algebraic principles, tackling complex mathematical problems becomes more manageable. Practicing regularly with algebraic equations strengthens one's skills in mathematical reasoning and problem-solving.