Ch. 1 - Equations and Inequalities

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

Lial 13th Edition

Ch. 1 - Equations and Inequalities

Problem 46

Chapter 2, Problem 46

Solve each equation or inequality. | 4 - 4x | + 2 = 4

Verified step by step guidance

Start by isolating the absolute value expression. Subtract 2 from both sides of the equation to get: \(|4 - 4x| = 4 - 2\).

Simplify the right side to find the value inside the absolute value equals: \(|4 - 4x| = 2\).

Recall that if \(|A| = B\), then \(A = B\) or \(A = -B\). Apply this to get two separate equations: \(4 - 4x = 2\) and \(4 - 4x = -2\).

Solve each equation for \(x\) separately. For \(4 - 4x = 2\), subtract 4 from both sides and then divide by -4. For \(4 - 4x = -2\), do the same.

Write the solutions from both equations as the solution set to the original equation.

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Key Concepts

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

Absolute Value Equations

An absolute value equation involves expressions within absolute value bars, which represent the distance from zero on the number line. To solve, isolate the absolute value expression and then set up two separate equations: one where the expression equals the positive value, and one where it equals the negative value.

Isolating the Absolute Value Expression

Before solving an absolute value equation, you must isolate the absolute value term on one side of the equation. This often involves performing inverse operations such as addition, subtraction, multiplication, or division to simplify the equation and prepare it for splitting into two cases.

Solving Linear Equations

After splitting the absolute value equation into two linear equations, solve each by isolating the variable using inverse operations. This includes combining like terms and dividing or multiplying to find the value of the variable that satisfies each equation.

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Height of a Projectile A projectile is launched from ground level with an initial velocity of v₀ feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t²+v₀t. In each exercise, find the time(s) that the projectile will (a) reach a height of 80 ft and (b) return to the ground for the given value of v₀. Round answers to the nearest hundredth if necessary. v₀=32

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