Problem 10 - A Graphical Approach to College Algebra

Question

Solve each equation by hand. Do not use a calculator. $$x-5=\sqrt{5 x-1}$$

Step-by-Step Solution

Answer

The solution is \(x = 13\).

Step 1: Isolate the Square Root

Start by isolating the square root expression on one side of the equation. Our equation is \( x - 5 = \sqrt{5x - 1} \). The square root is already isolated on the right side.

Step 2: Square Both Sides

Eliminate the square root by squaring both sides of the equation. This gives us: \((x - 5)^2 = (\sqrt{5x - 1})^2\).The equation becomes: \((x - 5)^2 = 5x - 1\).

Step 3: Expand the Binomial Expression

Expand the left side using the binomial expansion formula \((a-b)^2 = a^2 - 2ab + b^2\):\((x - 5)^2 = x^2 - 10x + 25\).Substitute this back into the equation, \(x^2 - 10x + 25 = 5x - 1\).

Step 4: Rearrange Terms to Form a Quadratic Equation

Bring all terms to one side to set up the quadratic equation: \(x^2 - 10x + 25 - 5x + 1 = 0\).Simplify to obtain: \(x^2 - 15x + 26 = 0\).

Step 5: Solve the Quadratic Equation

We can solve \(x^2 - 15x + 26 = 0\) using the quadratic formula,\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -15\), and \(c = 26\).Substitute these values into the formula:\(x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 26}}{2 \cdot 1}\).This simplifies to:\(x = \frac{15 \pm \sqrt{225 - 104}}{2}\).\(x = \frac{15 \pm \sqrt{121}}{2}\).\(x = \frac{15 \pm 11}{2}\).

Step 6: Calculate Possible Solutions for x

Calculate the two possible solutions:1. \(x = \frac{15 + 11}{2} = \frac{26}{2} = 13\).2. \(x = \frac{15 - 11}{2} = \frac{4}{2} = 2\).

Step 7: Verify Solutions

Check whether each solution satisfies the original equation:1. For \(x = 13\), substitute back: \(13 - 5 = \sqrt{5 \times 13 - 1}\) which simplifies to \(8 = \sqrt{64}\), thus true.2. For \(x = 2\), substitute back: \(2 - 5 = \sqrt{5 \times 2 - 1}\) which simplifies to \(-3 = \sqrt{9}\), not true.Therefore, \(x = 13\) is the only valid solution.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in algebra. It's used for solving quadratic equations, which are polynomial equations of the form \( ax^2 + bx + c = 0 \). A quadratic equation might look intimidating because of its squared term, but the quadratic formula breaks it down into simple steps. Here's the formula:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this formula:

• "\( a \)" is the coefficient of \( x^2 \).
• "\( b \)" is the coefficient of \( x \).
• "\( c \)" is the constant term.

The part under the square root, \( b^2 - 4ac \), is known as the discriminant. It tells us about the nature of the roots:

• If the discriminant is positive, there are two real and distinct solutions.
• If it is zero, there is exactly one real solution, known as a repeated root.
• If it is negative, there are no real solutions, but two complex solutions.

When you substitute \( a \), \( b \), and \( c \) from your equation into the formula, you can solve for \( x \) easily. It's powerful because it works for any quadratic equation.

Binomial Expansion

Binomial expansion is a method used to expand expressions raised to a power, such as \((a + b)^2\). It allows us to simplify and solve equations where terms are added or subtracted and then squared. The general rule for expanding \((a - b)^2\) is:

• \( (a - b)^2 = a^2 - 2ab + b^2 \)

For example, in the expression \((x - 5)^2\), we'd:

• Square the first term: \( x^2 \)
• Double the product of the two terms: \(-2 \cdot x \cdot 5 = -10x \)
• Square the second term: \( 25 \)
• Add these results together: \( x^2 - 10x + 25 \)

This method doesn't only apply to squares. It is part of the bigger binomial theorem used in extensive algebraic applications. But at its core, binomial expansion helps break down complex expressions into more manageable parts, making it easier to manipulate and solve.

Square Roots

Square roots are the inverse operation of squaring a number. For example, the square root of 9 is 3 because \( 3^2 = 9 \). They play a critical role in solving equations, especially when dealing with squares, as they "undo" the squaring. When you have a square root in an equation, it's often helpful to square both sides to eliminate it. This step was taken in the original exercise:

• By squaring \( x - 5 = \sqrt{5x - 1} \), it transforms into \( (x - 5)^2 = 5x - 1 \), simplifying the equation to a more solvable form.

Be cautious when working with square roots, as they can lead to extraneous solutions—solutions that do not satisfy the original equation. This occurs because squaring can mask the actual signs of numbers. Therefore, it's crucial to verify your solutions by substituting them back into the original equation. In our problem, only one solution was valid upon verification. Understanding square roots, therefore, is an essential skill in mastering quadratic equations and ensuring that solutions are correct.