Problem 10 - Algebra 1: Concepts and Skills

Question

Solve the quadratic equation by completing the square. $$ x^{2}-2 x-18=0 $$

Step-by-Step Solution

Answer

The solutions for the quadratic equation are \( x = 1 + \sqrt{19} \) and \( x = 1 - \sqrt{19} \).

Step 1: Adding 18 to both sides

Rearrange the equation by moving the constant term to the right side of the equation. Therefore, \(x^{2} -2x = 18 \).

Step 2: Completing the square

Complete the square by adding the square of half the coefficient of \( x \) on the both sides. So it becomes \( x^2 - 2x + 1 = 18 + 1\), which is \( (x-1)^{2} =19 \).

Step 3: Solve for \( x \)

Now, use the square root property \( (x)^2 = a \) to solve for \( x \). Then, \( x-1 = ± \sqrt{19} \), gives \( x = 1 ± \sqrt{19} \).

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations that involves altering the equation into a perfect square trinomial. This technique can be particularly useful when a quadratic equation doesn’t factor easily or when using the quadratic formula seems too cumbersome.

To begin with, we work with the equation \( x^{2} - 2x = 18 \). The goal is to turn the left-hand side of the equation into a perfect square trinomial, which is an expression that can be written as \((x - p)^{2} \), where \(p\) is a constant. To do this, we need to add \( (\frac{b}{2})^{2} \) to both sides of the equation, where \(b\) is the coefficient of \(x\) in the original quadratic equation. In our case, \(b = -2\), so we add \( (\frac{-2}{2})^{2} = 1 \) to each side, resulting in a new equation: \( x^{2} - 2x + 1 = 19 \).

The expression \( x^{2} - 2x + 1 \) is now a perfect square trinomial because it can be factored into \((x - 1)^{2}\), which looks much neater and is easier to solve. By working out the square root of both sides in the next step, we progress towards finding the value of \(x\).

Square Root Property

The square root property is crucial for solving equations where the variable appears within a square, such as \((x - p)^{2} = q\), where \(q\) is a nonnegative number. According to this property, if \( a^{2} = b \), then \( a = ± \sqrt{b}\), which simply means that if the square of \(a\) equals \(b\), then \(a\) is equal to either the positive or negative square root of \(b\).

In our problem, after completing the square, we have the equation \((x - 1)^{2} = 19\). To find the values of \(x\), we apply the square root property, taking the square root of both sides to get \(x - 1 = ± \sqrt{19}\). This gives us two possible solutions for \(x\) when we move \(1\) to the right-hand side, resulting in \(x = 1 ± \sqrt{19}\). Since \( \sqrt{19} \) is an irrational number, our final solutions are in this form rather than as exact numbers.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^{2} + bx + c = 0\). The formula is \( x = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a} \), where \(a\), \(b\), and \(c\) are the coefficients from the original equation. You can use this formula when completing the square seems too tedious or factoring is not possible.

In the context of our example, though we've earlier solved the equation by completing the square, you might be curious to see how it would work using the quadratic formula. With \(a = 1\), \(b = -2\), and \(c = -18\), the formula would yield the same solutions as before: \(x = \frac{-(-2) ± \sqrt{(-2)^{2} - 4*1*(-18)}}{2*1} \), simplifying to \(x = \frac{2 ± \sqrt{4 + 72}}{2} \), and eventually to \(x = 1 ± \sqrt{19}\).

Each method—completing the square, utilizing the square root property, and the quadratic formula—offers different advantages depending on the specific quadratic equation you're working with.