Copyright © Cengage Learning. All rights reserved. 1.4 QUADRATIC EQUATIONS AND APPLICATIONS Copyright © Cengage Learning. All rights reserved.

What You Should Learn • Solve quadratic equations by factoring. • Solve quadratic equations by extracting square roots. • Solve quadratic equations by completing the square. • Use the Quadratic Formula to solve quadratic equations. • Use quadratic equations to model and solve real-life problems.

Factoring

Example 1(a) – Solving a Quadratic Equation by Factoring 2x2 + 9x + 7 = 3 2x2 + 9x + 4 = 0 (2x + 1)(x + 4) = 0 2x + 1 = 0 x = x + 4 = 0 x = –4 The solutions are x = and x = –4. Check these in the original equation. Original equation Write in general form. Factor. Set 1st factor equal to 0. Set 2nd factor equal to 0.

Example 1(b) – Solving a Quadratic Equation by Factoring cont’d 6x2 – 3x = 0 3x(2x – 1) = 0 3x = 0 x = 0 2x – 1 = 0 x = The solutions are x = 0 and x = . Check these in the original equation. Original equation Factor. Set 1st factor equal to 0. Set 2nd factor equal to 0.

Extracting Square Roots

Example 2 – Extracting Square Roots Solve each equation by extracting square roots. a. 4x2 = 12 b. (x – 3)2 = 7 Solution: a. 4x2 = 12 x2 = 3 x =  When you take the square root of a variable expression, you must account for both positive and negative solutions. Write original equation. Divide each side by 4. Extract square roots.

Example 2 – Solution cont’d So, the solutions are x = and x = – . Check these in the original equation. b. (x – 3)2 = 7 x – 3 =  x = 3  The solutions are x = 3  . Check these in the original equation. Write original equation. Extract square roots. Add 3 to each side.

Completing the Square

Example 3 – Completing the Square: Leading Coefficient Is 1 Solve x2 + 2x – 6 = 0 by completing the square. Solution: x2 + 2x – 6 = 0 x2 + 2x = 6 x2 + 2x + 12 = 6 + 12 Write original equation. Add 6 to each side. Add 12 to each side.

Example 3 – Solution (x + 1)2 = 7 x + 1 =  x = –1  cont’d (x + 1)2 = 7 x + 1 =  x = –1  The solutions are x = –1  . Check these in the original equation as follows. Simplify. Take square root of each side. Subtract 1 from each side.

Example 3 – Solution Check: x2 + 2x – 6 = 0 cont’d Check: x2 + 2x – 6 = 0 (–1 + )2 + 2 (–1 + ) – 6 ≟ 0 8 – 2 – 2 + 2 – 6 ≟ 0 8 – 2 – 6 = 0 Check the second solution in the original equation. Write original equation. Substitute –1 + for x. Multiply. Solution checks.

The Quadratic Formula

The Quadratic Formula The Quadratic Formula is one of the most important formulas in algebra. You should learn the verbal statement of the Quadratic Formula: “Negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.”

The Quadratic Formula In the Quadratic Formula, the quantity under the radical sign, b2 – 4ac, is called the discriminant of the quadratic expression ax2 + bx + c. It can be used to determine the nature of the solutions of a quadratic equation.

Example 6 – The Quadratic Formula: Two Distinct Solutions Use the Quadratic Formula to solve x2 + 3x = 9. Solution: The general form of the equation is x2 + 3x – 9 = 0. The discriminant is b2 – 4ac = 9 + 36 = 45, which is positive. So, the equation has two real solutions. You can solve the equation as follows. x2 + 3x – 9 = 0 Write in general form. Quadratic Formula

Example 6 – Solution The two solutions are: cont’d The two solutions are: Check these in the original equation. Substitute a = 1, b = 3, and c = –9. Simplify. Simplify.

Applications

Example 7 – Finding the Dimensions of a Room A bedroom is 3 feet longer than it is wide (see Figure 1.20) and has an area of 154 square feet. Find the dimensions of the room. Figure 1.20

Example 7 – Solution Verbal Model: Labels: Width of room = w (feet) Length of room = w + 3 (feet) Area of room = 154 (square feet) Equation: w(w + 3) = 154 w2 + 3w – 154 = 0 (w – 11)(w + 14) = 0

Example 7 – Solution w – 11 = 0 w = 11 w + 14 = 0 w = –14 cont’d w – 11 = 0 w = 11 w + 14 = 0 w = –14 Choosing the positive value, you find that the width is 11 feet and the length is w + 3, or 14 feet. You can check this solution by observing that the length is 3 feet longer than the width and that the product of the length and width is 154 square feet.

Applications Another common application of quadratic equations involves an object that is falling (or projected into the air). The general equation that gives the height of such an object is called a position equation, and on Earth’s surface it has the form s = –16t2 + v0t + s0. In this equation, s represents the height of the object (in feet), v0 represents the initial velocity of the object (in feet per second), s0 represents the initial height of the object (in feet), and t represents the time (in seconds).

Applications A third type of application of a quadratic equation is one in which a quantity is changing over time t according to a quadratic model. A fourth type of application that often involves a quadratic equation is one dealing with the hypotenuse of a right triangle. In these types of applications, the Pythagorean Theorem is often used.

Applications The Pythagorean Theorem states that a2 + b2 = c2 where a and b are the legs of a right triangle and c is the hypotenuse. Pythagorean Theorem