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Chapter 1: Problem 48

Solve equation by completing the square. $$ x^{2}+6 x=-8 $$

Short Answer

Expert verified
The solutions for the equation are \(x = -3 + \sqrt{17}\) and \(x = -3 - \sqrt{17}\).

Step by step solution

01

Isolate the square terms

Firstly, separate the square terms and the constant on opposite sides of the equation. Rearrange \(x^{2}+6 x=-8\) to \(x^{2}+6 x=8\) by adding 8 to both sides.
02

Complete the square

In order to complete the square, add the square of half the coefficient of \(x\) on both sides. The coefficient of \(x\) is 6, hence half of it would be 3. The square of 3 is 9. Thus the equation becomes \(x^{2}+6 x + 9 = 8 + 9\), which simplifies to \(x^{2}+6 x + 9 = 17\). The left side of the equation can now be expressed as a perfect square and the equation becomes \((x+3)^2 = 17\).
03

Solve for \(x\)

Finally, solve for \(x\) by square rooting both sides of the equation. This visualizes as \(x+3 = 卤 \sqrt{17}\). Thus, \(x = -3 卤 \sqrt{17}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are equations where the highest degree of the variable is two. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\). Quadratics can graph into a parabola and have various methods of solutions such as factoring, using the quadratic formula, or completing the square.
Understanding the structure helps in identifying the equation type and choosing the method to solve accordingly. In our example, \(x^2 + 6x = -8\), the equation needs transformation to isolate terms and solve.
Solving Equations
Solving equations involves finding values for the variable that make the equation true. For quadratic equations, the solutions can provide the x-intercepts of the parabola if the equation is set to zero.
To solve \(x^2 + 6x = -8\), it's important to first rearrange terms to prepare for completing the square. This process assists in finding exact solutions and understanding variable behavior.
Perfect Square Trinomial
A perfect square trinomial is an expression like \((x + a)^2\), which can be expanded to \(x^2 + 2ax + a^2\). Completing the square uses this concept to transform a quadratic into a trinomial square that mirrors this structure.
In the exercise, \(x^2 + 6x\) was rearranged to include \(9\) (half of \(6\), squared), turning it into \((x + 3)^2 = 17\). Recognizing this trinomial helps simplify solving equations by reducing them to a familiar format.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions using arithmetic operations. It's a foundational skill for solving equations, isolating variables, and transforming expressions.
In the given problem, manipulation began by adding \(8\) to both sides of the equation to isolate the quadratic and linear terms. Then, additional manipulation was applied to complete the square by introducing \(9\) to balance the equation. Finally, solving required square-rooting both sides and performing basic operations to express \(x\) in terms of known values.

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Most popular questions from this chapter

In Exercises 59鈥�94, solve each absolute value inequality. $$ \left|2-\frac{x}{2}\right|-1 \leq 1 $$

Find all values of \(x\) satisfying the given conditions. $$ y_{1}=\frac{3}{x-1}, y_{2}=\frac{8}{x}, \text { and } y_{1}+y_{2}=3 $$

When the sum of 6 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

In a round-robin chess tournament, each player is paired with every other player once. The formula $$ N=\frac{x^{2}-x}{2} $$ models the number of chess games, \(N,\) that must be played in a round-robin tournament with \(x\) chess players. Use this formula to solve. In a round-robin chess tournament, 21 games were played. How many players were entered in the tournament?

Find all values of \(x\) satisfying the given conditions. $$ \begin{aligned} &y_{1}=2 x^{2}+5 x-4, y_{2}=-x^{2}+15 x-10, \text { and } &y_{1}-y_{2}=0 \end{aligned} $$
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