Problem 51 - Algebra and Trigonometry

Question

Solve equation by completing the square. $$ x^{2}-6 x-11=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 3 + 2\sqrt{5}\) and \(x = 3 - 2\sqrt{5}\).

Step 1: Rearrange the Equation

Rearrange the equation in the form of \(x^2 - bx = c\). This gives us the equation \(x^2 - 6x = 11\).

Step 2: Completing the Square

To complete the square for the equation \(x^2 - 6x = 11\), we need to add \((\frac{b}{2})^2\) to both sides of the equation. In this case, \(b\) is equal to -6. Calculating \((\frac{-6}{2})^2\), we get 9. So, we add 9 to both sides resulting in the equation \((x - 3)^2 = 20\).

Step 3: Solve for x

Now take square roots on both sides of the equation. This gives two solutions: \(x = 3 + \sqrt{20}\) and \(x = 3 - \sqrt{20}\).

Step 4: Simplify \(\sqrt{20}\)

Simplify \(\sqrt{20}\) to \(2\sqrt{5}\) to get the final solutions: \(x = 3 + 2\sqrt{5}\) and \(x = 3 - 2\sqrt{5}\).

Key Concepts

Quadratic Equations

Quadratic equations form a fundamental part of algebra and are expressed in the general form \(ax^2 + bx + c = 0\). These equations involve a variable raised to the power of two, referred to as a square, hence the term "quadratic."
The first key concept of a quadratic equation is that its graph forms a parabola, which can either open upwards or downwards depending on the sign of the coefficient \(a\). For example:

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

Completing the square is one way to solve quadratic equations. It transforms the equation into a form that makes it easy to find the values of \(x\). It's particularly useful when the equation doesn't factor neatly, allowing us to find the roots or solutions of the quadratic without directly solving it.
In the given exercise, the quadratic equation \(x^2 - 6x - 11 = 0\) needs to be solved using this method. Completing the square involves creating a perfect square trinomial on one side to easily obtain the solutions.

Square Root

The square root is a concept integral to solving quadratic equations, especially when using the method of completing the square. A square root of a number \(a\) is a value \(b\) such that \(b^2 = a\). This concept is represented by the symbol \(\sqrt{}\).
When you "take the square root" in the context of completing the square, you're effectively determining two possible values: one positive and one negative. This duality arises because both \(b\) and \(-b\) satisfy \(b^2 = a\). In the example provided, after completing the square, we have \((x-3)^2 = 20\). To solve for \(x\), we take the square root of both sides, leading to \(x - 3 = \pm \sqrt{20}\). This results in two expressions for \(x\):

• \(x = 3 + \sqrt{20}\)
• \(x = 3 - \sqrt{20}\)

Knowing how to handle square roots, such as simplifying \(\sqrt{20}\), is crucial for finding these solutions precisely. The goal is always to express it in its simplest terms to provide clear and concise answers.

Simplification

Simplification is an important step in mathematical problem-solving, including working with quadratic equations.
After taking square roots, the solutions often appear with radicals that may not be in their simplest form. Simplifying these radicals can provide cleaner and more interpretable solutions. For example, once we have \(x = 3 + \sqrt{20}\) and \(x = 3 - \sqrt{20}\), we should simplify \(\sqrt{20}\). To do this, we find factors of 20 that are perfect squares, which in this case is 4. Thus, \(\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \cdot \sqrt{5} = 2\sqrt{5}\). This simplification leads us to the final solutions:

• \(x = 3 + 2\sqrt{5}\)
• \(x = 3 - 2\sqrt{5}\)

Simplifying radicals helps to keep our answers tidy and easy to understand. It ensures the beauty and power of mathematics shine through, making problem-solving a satisfying experience.