Problem 74 - Elementary Algebra

Question

For the following problems, solve the equations. $$ \sqrt{4 x-8}=x-2 $$

Step-by-Step Solution

Answer

Answer: The values of x that satisfy the equation are x = 2 and x = 6.

Step 1: Isolate the square root

The given equation is: $$ \sqrt{4x-8} = x-2 $$ The square root is already isolated, so we can move to the next step.

Step 2: Square both sides of the equation

To get rid of the square root, we will square both sides of the equation: $$ \left(\sqrt{4x-8}\right)^2 = (x-2)^2 $$

Step 3: Simplify the equation

Now we can simplify the equation by applying the squared operation to both sides: $$ 4x-8 = (x-2)^2 $$ Expand \((x-2)^2\): $$ 4x-8 = x^2 - 4x + 4 $$

Step 4: Solve the quadratic equation

Move all terms to the right side of the equation to form a quadratic equation: $$ 0 = x^2 - 8x + 12 $$ Now we need to either find the roots (x-intercepts) by factorizing or use the quadratic equation formula. Let's try factoring: $$ 0 = (x - 6)(x - 2) $$ So, we have 2 possible solutions for x: $$ x - 6 = 0 \Rightarrow x = 6 \\ x - 2 = 0 \Rightarrow x = 2 $$

Step 5: Check the solutions

Sometimes, when solving square root equations, we may find extraneous solutions (solutions that don't work in the original equation). Let's check if both solutions are valid: For \(x=6\): $$ \sqrt{4(6)-8} = 6 - 2 \\ \sqrt{16}=4 $$ This solution is valid. For \(x=2\): $$ \sqrt{4(2)-8} = 2 - 2 \\ 0 = 0 $$ This solution is also valid.

Step 6: Write the final solution

Since both solutions are valid, the final solution is: $$ x = 2, 6 $$

Key Concepts

square root equations

Square root equations are mathematical equations that contain a variable inside a square root. Solving these requires isolating the square root expression and then squaring both sides of the equation to eliminate the square root. The exercise provided was:
\(\sqrt{4x - 8} = x - 2\).

• The square root is already isolated on the left side of the equation.
• To solve, square both sides so the equation becomes \((\sqrt{4x - 8})^2 = (x - 2)^2\).

This transforms the equation into one we can simplify further. It's crucial to isolate the square root at the start to make it easier to eliminate by squaring. After removing the square root, we shift our focus to solving the resulting equation.

factoring quadratics

Factoring quadratics is a common method used to solve quadratic equations. Quadratic equations take the general form \(ax^2 + bx + c = 0\). In simple cases, you can factor the quadratic equation into the product of binomials. For example, the equation given after simplifying:
\(0 = x^2 - 8x + 12\)
can be factored into two binomials. Here's how it works:

• Look for two numbers that multiply to \(+12\) (the constant term) and add to \(-8\) (the coefficient of x).
• These numbers are \(-6\) and \(-2\), factoring gives: \((x - 6)(x - 2) = 0\).

Factoring works well when the quadratic can be easily broken down into simpler expressions. Solving each binomial \((x - 6) = 0\) or \((x - 2) = 0\) provides the solutions \(x = 6\) and \(x = 2\). Factoring is often the fastest way once you are familiar with spotting factorable quadratics.

extraneous solutions

Extraneous solutions are potential solutions that arise during the calculation process but do not satisfy the original equation. When dealing with square roots, it's important to check each potential solution by plugging it back into the original equation.
In this exercise, by squaring both sides, there was a need to verify if both solutions actually work in the original equation:

• For \(x=6\), substitute back into the original: \(\sqrt{4(6)-8} = 6 - 2\) which simplifies to \(4 = 4\), validating the solution.
• For \(x=2\), substitute: \(\sqrt{4(2)-8} = 2 - 2\) simplifies to \(0 = 0\), confirming this solution as well.

A solution that does not satisfy the original equation after substitution is extraneous. Always check your answers, particularly when dealing with square root equations, to ensure they are valid solutions.

quadratic formula

The quadratic formula provides a guaranteed method to find the solutions to any quadratic equation \(ax^2 + bx + c = 0\). This formula is especially useful when a quadratic equation does not factor neatly. The formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the exercise, although factoring was straightforward, the quadratic formula could be applied for confirmation or when factoring proves difficult.

• Substitute the coefficients \(a = 1\), \(b = -8\), and \(c = 12\) from \(x^2 - 8x + 12 = 0\) into the quadratic formula.
• Calculate the discriminant \(b^2 - 4ac\); here it equals 4, greater than zero, indicating two distinct real solutions.
• Complete the formula to find \(x = 6\) and \(x = 2\), as already determined by factoring.

The quadratic formula is universal and helpful especially when factorization isn’t possible or apparent. Understanding its use is crucial for mastering quadratic equations.