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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblem 49
Chapter 2, Problem 49

Solve each equation in Exercises 47–64 by completing the square. x2−2x=2x^2 - 2x = 2

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1
Start with the given quadratic equation: \(x^2 - 2x = 2\).
To complete the square, first move the constant term to the right side (if it's not already isolated): \(x^2 - 2x = 2\) (already isolated in this case).
Take half of the coefficient of \(x\), which is \(-2\), divide by 2 to get \(-1\), then square it to get \((-1)^2 = 1\).
Add this square (1) to both sides of the equation to maintain equality: \(x^2 - 2x + 1 = 2 + 1\).
Rewrite the left side as a perfect square trinomial: \((x - 1)^2 = 3\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Completing the Square

Completing the square is a method used to solve quadratic equations by transforming the equation into a perfect square trinomial. This involves adding a specific value to both sides of the equation to create a binomial squared, making it easier to solve for the variable.
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Quadratic Equations

A quadratic equation is a second-degree polynomial equation in the form ax² + bx + c = 0. Understanding its structure is essential because methods like completing the square rely on manipulating the equation to isolate the variable and find its roots.
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Isolating the Variable

Isolating the variable means rearranging the equation so that the variable term stands alone on one side. This step is crucial before completing the square, as it allows you to clearly add the necessary constant to both sides and solve for the variable.
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Equations with Two Variables
Related Practice
Textbook Question

Exercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 1/(x - 1) + 5 = 11/(x - 1)

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Textbook Question

Write each English sentence as an equation in two variables. Then graph the equation. The y-value is three decreased by the square of the x-value.

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Textbook Question

In Exercises 35–54, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? S = C/(1 - r) for r

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Textbook Question

Write each English sentence as an equation in two variables. Then graph the equation. The y-value is the difference between four and twice the x-value.

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Textbook Question

In all exercises, other than exercises with no solution, use interval notation to express solution sets and graph each solution set on a number line. In Exercises 27–50, solve each linear inequality. 5(x - 2) - 3(x + 4) ≥ 2x - 20

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Textbook Question

Perform the indicated operations and write the result in standard form. −8(−3−−5)\(\sqrt{-8}\) \(\left\)( \(\sqrt{-3}\) - \(\sqrt{-5}\) \(\right\))

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