Problem 9 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation by hand. Do not use a calculator. $$x-4=\sqrt{3 x-8}$$

Step-by-Step Solution

Answer

The solution is \(x = 8\).

Step 1: Isolate the Square Root

Start by isolating the square root on one side of the equation. Since the equation is \(x - 4 = \sqrt{3x - 8}\), the square root is already isolated.

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation: \((x - 4)^2 = (\sqrt{3x - 8})^2\). This simplifies to \(x^2 - 8x + 16 = 3x - 8\).

Step 3: Move All Terms to One Side

Rearrange the equation so that all terms are on one side: \(x^2 - 8x + 16 - 3x + 8 = 0\). Simplify to get \(x^2 - 11x + 24 = 0\).

Step 4: Solve the Quadratic Equation by Factoring

Factor the quadratic equation \(x^2 - 11x + 24 = 0\). Look for two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8, so the equation factors to \((x - 3)(x - 8) = 0\).

Step 5: Set Each Factor to Zero and Solve

Set each factor equal to zero: \(x - 3 = 0\) or \(x - 8 = 0\). Solving these gives \(x = 3\) and \(x = 8\).

Step 6: Verify Solutions

Substitute \(x = 3\) back into the original equation: \(3 - 4 = \sqrt{3 \times 3 - 8}\), which simplifies to \(-1 = 1\), a false statement. Now substitute \(x = 8\): \(8 - 4 = \sqrt{3 \times 8 - 8}\), which simplifies to \(4 = 4\), a true statement. Only \(x = 8\) is a valid solution.

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra and is typically expressed in the form \( ax^2 + bx + c = 0 \). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). These equations are named "quadratic" because they involve the variable raised to the second power, or squared. Quadratic equations can have zero, one, or two real solutions depending on the values of the coefficients and the discriminant, which is \(b^2 - 4ac\).

The solution of a quadratic equation can usually be found using various methods, such as:

• Factoring: Expressing the quadratic equation as a product of its linear factors if possible.
• Quadratic formula: Solving using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the square: Rewriting the equation to make it easier to solve.

In the exercise given, the quadratic equation formed is \(x^2 - 11x + 24 = 0\). This equation was solved by factoring, which is often the simplest method when the equation can be easily factored.

Square Root Isolation

Square root isolation is a technique used to simplify equations involving square roots. The main idea is to rearrange the equation so that the square root is by itself on one side. Doing so allows us to eliminate the square root by squaring both sides of the equation.

Let's take a look at the original equation from the exercise: \(x - 4 = \sqrt{3x - 8}\). In this case, the square root \(\sqrt{3x - 8}\) is already isolated on the right side of the equation. This means we can proceed to the next step, which is to square both sides of the equation to remove the square root.

• Square both sides: \((x - 4)^2 = (\sqrt{3x - 8})^2\), simplifying to \(x^2 - 8x + 16 = 3x - 8\).

This step allows us to transform the equation into a quadratic form that can be solved using typical algebraic methods.

Factoring

Factoring is a method often used to solve quadratic equations by breaking them down into simpler expressions. If a quadratic equation can be written as the product of two binomials, it can be solved quite easily by setting each factor equal to zero. This approach can efficiently find the roots or solutions of the equation.

In the exercise, we first reformulated the result after isolating the square root and squaring both sides into \(x^2 - 11x + 24 = 0\). To factor this quadratic equation:

• Look for two numbers that multiply to 24 (the last term) and add up to -11 (the middle term). In this case, those numbers are -3 and -8.
• Rewrite the equation as \((x - 3)(x - 8) = 0\).

The solutions are found by setting each factor equal to zero:

• \(x - 3 = 0\) gives \(x = 3\).
• \(x - 8 = 0\) gives \(x = 8\).

However, after substituting these back into the original equation, only \(x = 8\) satisfies it, meaning \(x = 8\) is the valid solution.