Problem 14 - Algebra 2

Question

Solve each equation by completing the square. \(x^{2}+2 x+6=0\)

Step-by-Step Solution

Answer

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

Step 1: Rearrange the Equation

First, move the constant term to the other side of the equation:\[ x^2 + 2x = -6 \]

Step 2: Find the Term to Complete the Square

To complete the square, we need to add a specific value to both sides of the equation. This value is found by taking half of the coefficient of \(x\), squaring it, and adding it to both sides. The coefficient of \(x\) is 2, so half of it is 1, and squaring it gives 1.Add 1 to both sides:\[ x^2 + 2x + 1 = -6 + 1 \]

Step 3: Write the Perfect Square Trinomial

The left side of the equation is now a perfect square trinomial:\[ (x + 1)^2 = -5 \]

Step 4: Solve for x by Taking Square Roots

Take the square root of both sides, remembering that there are positive and negative square roots:\[ x + 1 = \pm \sqrt{-5} \]Since the square root of a negative number involves an imaginary number, represent it as:\[ x + 1 = \pm i\sqrt{5} \]

Step 5: Isolate x

Subtract 1 from both sides to solve for \(x\):\[ x = -1 \pm i\sqrt{5} \]

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two, usually written in the form \(ax^2 + bx + c = 0\), where *a*, *b*, and *c* are constants, and \(x\) represents an unknown variable. Quadratics are fundamental in algebra and can describe various practical scenarios, including projectile motion and real-world measurements.

• The equation \(x^2 + 2x + 6 = 0\) is a quadratic with coefficients *a* = 1, *b* = 2, and *c* = 6.
• To solve quadratics, methods like factoring, using the quadratic formula, and completing the square are common.

Completing the Square is a helpful technique, especially when the quadratic does not easily factorize. It involves altering the equation into a perfect square trinomial, which simplifies finding solutions.

Imaginary Numbers

Imaginary numbers arise when solving equations that involve taking the square root of negative numbers. The imaginary unit is denoted by \(i\), defined as \(i = \sqrt{-1}\). Using \(i\), we can express square roots of negative numbers in terms of real numbers.

• For our example, after completing the square, we get the expression \( (x + 1)^2 = -5\).
• Taking the square root on both sides introduces the imaginary number \(i\), giving \(\pm i\sqrt{5}\).

Imaginary numbers are essential in complex analysis and many areas of engineering and physics. Understanding them broadens our ability to address problems involving non-real solutions.

Perfect Square Trinomial

A perfect square trinomial is an expression that can be written as the square of a binomial. For example, \((x + 1)^2\) expands to \(x^2 + 2x + 1\). This form is significant because it allows us to solve equations by reversing the process — compressing a trinomial into a square of a binomial.

• The standard method to identify or create a perfect square trinomial is adding the square of half the linear term's coefficient to the equation.
• For instance, in \(x^2 + 2x + 6 = 0\), we added \(1\) to reach \((x + 1)^2\).
• This conversion creates a simple equation where solving can proceed easily.

Understanding how to create and identify these trinomials accelerates solving quadratic equations, making them less complex and more intuitive.

Solving Equations

Solving equations is a key operation in algebra, involving finding the value(s) of variables that make an equation true. For quadratics, we aim to find the \(x\) values that satisfy \(ax^2 + bx + c = 0\). Various techniques exist, each useful in distinct scenarios. Completing the square is one such method. It allows transformation into a form that is easier to manipulate.

• Once a perfect square trinomial is formed, take the square root of both sides of the equation.
• Remember to consider both the positive and negative roots: \(x + 1 = \pm i\sqrt{5}\).
• Finally, isolate \(x\) to determine the solutions: \(x = -1 \pm i\sqrt{5}\).

This method reveals both real and imaginary solutions, illustrating the power of algebraic techniques in uncovering the full range of solutions for an equation.