Problem 39 - Algebra A Combined Function

Question

Solve each equation by completing the square. $$ x^{2}+x-1=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).

Step 1: Move constant to the right side

The given equation is \(x^2 + x - 1 = 0\). First, we move the constant term to the other side of the equation to focus on the quadratic and linear terms: \(x^2 + x = 1\).

Step 2: Find the value to complete the square

To complete the square, we need to add a specific value to both sides of the equation. This value is calculated by taking half of the coefficient of \(x\), squaring it, and adding it to both sides. Here, the coefficient of \(x\) is 1. Half of 1 is \(\frac{1}{2}\), and squaring this gives \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

Step 3: Add \(\frac{1}{4}\) to both sides

Add \(\frac{1}{4}\) to both sides of the equation: \(x^2 + x + \frac{1}{4} = 1 + \frac{1}{4}\). Simplifying the right side gives \(x^2 + x + \frac{1}{4} = \frac{5}{4}\).

Step 4: Rewrite the left side as a binomial square

The left side \(x^2 + x + \frac{1}{4}\) is a perfect square trinomial and can be written as \((x + \frac{1}{2})^2\). Now our equation is \((x + \frac{1}{2})^2 = \frac{5}{4}\).

Step 5: Solve the equation using the square root property

To solve for \(x\), take the square root of both sides: \(x + \frac{1}{2} = \pm \sqrt{\frac{5}{4}}\). Simplify to get \(x + \frac{1}{2} = \pm \frac{\sqrt{5}}{2}\).

Step 6: Isolate x

Subtract \(\frac{1}{2}\) from both sides to isolate \(x\): \(x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2}\). This results in two solutions: \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).

Key Concepts

Quadratic Equation

Quadratic equations are a type of polynomial equation where the highest power of the variable, usually represented by \(x\), is two. The general form of a quadratic equation is:

• \(ax^2 + bx + c = 0\)

In this equation, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\) (as otherwise the equation would not be quadratic).
The task is to find the values of \(x\) that satisfy this equation, known as the roots or solutions.

Quadratic equations can be solved through several methods, such as factoring, using the quadratic formula, or completing the square. Completing the square is particularly useful when the equation does not factor neatly, as it transforms the equation into a perfect square trinomial, making it easier to solve.

Let's look at how "completing the square" works with the quadratic equation \(x^2 + x - 1 = 0\). So, we aim to manipulate the left-hand side into a neat square form, which then allows us to solve the equation easily.

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression that can be written as a square of a binomial. This means that:

• \(x^2 + 2ax + a^2 = (x + a)^2\)

In the exercise, we first focused on creating this form on the left side of the equation.
Starting with \(x^2 + x + \frac{1}{4}\), it was turned into a perfect square trinomial. The quadratic and linear terms were in place, and by adding \(\frac{1}{4}\), we achieved:

• \((x + \frac{1}{2})^2 = x^2 + x + \frac{1}{4}\)

Recognizing and forming a perfect square trinomial simplifies balancing and solving quadratic equations. It allows the simplifies application of the square root property in the next step.

Square Root Property

The square root property is a method used to solve equations that are set in the form of a perfect square. Once we rewrite a quadratic equation into the format \((x+a)^2 = b\), we can easily solve for \(x\) using this property.

The principle here is straightforward:

• If \((x+a)^2 = b\), then \(x+a = \pm \sqrt{b}\)

Taking square roots on both sides helps eliminate the square, simplifying the equation closer to its solution.
In our exercise, the perfect square was rewritten as \((x + \frac{1}{2})^2 = \frac{5}{4}\) and using the square root property:

• \(x + \frac{1}{2} = \pm \frac{\sqrt{5}}{2}\)

The ± symbol indicates that there are two potential solutions, reflecting the two points where the quadratic graph intersects the x-axis.
Finally, by isolating \(x\), we achieve the solutions:\(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\). The square root property thus forms the backbone of solving the newly established equation efficiently.