Problem 19 - Intermediate Algebra

Question

Use the method of completing the square to solve each quadratic equation. $$ y^{2}-10 y=1 $$

Step-by-Step Solution

Answer

The solutions are \( y = 5 + 2\sqrt{6} \) and \( y = 5 - 2\sqrt{6} \).

Step 1: Move the Constant to the Right Side

We start with the equation \( y^2 - 10y = 1 \). The first step is to move the constant term to the right side by subtracting 1 from both sides. This gives us: \[ y^2 - 10y = -1 \].

Step 2: Find the Value Needed to Complete the Square

To complete the square, we need to add and subtract a specific value on the left side. This value is found by taking half of the coefficient of \( y \), squaring it, and then adding it to the left side. Half of -10 is -5, and squaring -5 gives 25. So we need to add 25 to both sides to complete the square.

Step 3: Add and Subtract to Complete the Square

Add 25 to both sides of the equation. The equation \( y^2 - 10y = -1 \) becomes: \[ y^2 - 10y + 25 = 24 \].

Step 4: Rewrite as a Perfect Square Trinomial

The left side of the equation \( y^2 - 10y + 25 \) is now a perfect square trinomial. It can be rewritten as: \[ (y - 5)^2 = 24 \].

Step 5: Solve for y by Taking the Square Root

To find \( y \), take the square root of both sides of the equation: \( (y - 5)^2 = 24 \). This gives: \( y - 5 = \pm \sqrt{24} \). Simplifying \( \sqrt{24} \), we get \( 2\sqrt{6} \). Thus, \( y - 5 = \pm 2\sqrt{6} \).

Step 6: Isolate y

Add 5 to both sides to isolate \( y \): \( y = 5 \pm 2\sqrt{6} \). Therefore, the solutions are \( y = 5 + 2\sqrt{6} \) and \( y = 5 - 2\sqrt{6} \).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation with a degree of 2. This means the highest power of the variable in the equation is 2. In general, it looks like this: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations can appear in various forms, such as in our example equation, \( y^2 - 10y = 1 \).

To solve a quadratic equation, there are several methods available:

• The quadratic formula
• Factoring
• Graphing
• Completing the square

In this context, we focus on the method of completing the square, which involves making the left-hand side of the equation into a perfect square trinomial. This process enables us to easily solve the equation using the square root method. Understanding how quadratic equations work is essential as they often appear in real-world scenarios, from physics problems to financial calculations.

Perfect Square Trinomial

A perfect square trinomial is a special form of a trinomial (a polynomial with three terms) that can be expressed as the square of a binomial. It looks like this: \((x + a)^2 = x^2 + 2ax + a^2\).

The process of completing the square transforms a quadratic expression into this form. Let's apply this to the example \( y^2 - 10y \). We aim to rewrite it as a perfect square trinomial.

Here is the step-by-step way:

• Take the coefficient of \( y \), which is -10.
• Divide it by 2, giving -5.
• Square -5 to get 25.

By adding and subtracting this calculated value (25 in this case) to the left side of the equation, you create a perfect square trinomial:

\( y^2 - 10y + 25 = (y - 5)^2 \).

Turning your quadratic expression into a perfect square trinomial simplifies the process of solving the equation using the square root method.

Square Root Method

The square root method is a straightforward technique used to solve equations that are in the form of a perfect square. Once you have a perfect square trinomial, you can easily apply this method to find the solutions.

Consider the expression \((y - 5)^2 = 24\). To solve it using the square root method, follow these steps:

• Take the square root of both sides of the equation: \((y - 5)^2 = 24\)
• This results in two possible equations: \(y - 5 = \pm \sqrt{24}\)
• Simplifying \(\sqrt{24}\), it becomes \(2\sqrt{6}\)
• So, the equations become \(y - 5 = \pm 2\sqrt{6}\)

The last step is to isolate the variable \(y\) by adding 5 to both sides:

\(y = 5 \pm 2\sqrt{6}\).

This gives you the two solutions: \( y = 5 + 2\sqrt{6} \) and \( y = 5 - 2\sqrt{6} \). The square root method often provides an efficient means of finding solutions, especially after transforming the quadratic expression into a perfect square.