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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 95
Chapter 2, Problem 95

Solve each equation in Exercises 83–108 by the method of your choice. 3x2−12x+12=03x^2 - 12x + 12 = 0

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1
Identify the given quadratic equation: \(3x^2 - 12x + 12 = 0\).
Divide the entire equation by 3 to simplify it: \(x^2 - 4x + 4 = 0\).
Recognize that the simplified quadratic is a perfect square trinomial, which can be factored as \((x - 2)^2 = 0\).
Set the factor equal to zero: \(x - 2 = 0\).
Solve for \(x\) by adding 2 to both sides: \(x = 2\).

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Key Concepts

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

Quadratic Equations

A quadratic equation is a second-degree polynomial equation in the form ax² + bx + c = 0, where a ≠ 0. It represents a parabola when graphed and can have zero, one, or two real solutions depending on the coefficients.
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Introduction to Quadratic Equations

Factoring and Simplification

Factoring involves rewriting a quadratic equation as a product of simpler expressions to find its roots. Simplification may include dividing the entire equation by a common factor to make factoring or other methods easier.
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Quadratic Formula and Discriminant

The quadratic formula x = (-b ± √(b² - 4ac)) / 2a provides solutions to any quadratic equation. The discriminant (b² - 4ac) determines the nature of the roots: positive for two real roots, zero for one real root, and negative for complex roots.
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The Discriminant
Related Practice
Textbook Question

In Exercises 95–99, perform the indicated operations and write the result in standard form. 4/(2 + i)(3 - i)

Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. 9−6x+x2=09 - 6x + x^2 = 0

Textbook Question

In Exercises 95–102, use interval notation to represent all values of x satisfying the given conditions. y1=x2+3,y2=x3+52,andy1≤y2y_1 = \(\frac{x}{2}\) + 3, \(\quad\) y_2 = \(\frac{x}{3}\) + \(\frac{5}{2}\), \(\quad\) \(\text{and}\) \(\quad\) y_1 \(\leq\) y_2

Textbook Question

Evaluate x2 - x for the value of x satisfying 4(x - 2) + 2 = 4x - 2(2 - x).

Textbook Question

Solve the equations containing absolute value in Exercises 94–95. |2x+1| = 7

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Textbook Question

In Exercises 91–100, find all values of x satisfying the given conditions. y=x−x−2andy=4y = x - \(\sqrt{x - 2}\) \(\quad\) \(\text{and}\) \(\quad\) y = 4

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