2 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.
4 Example 1(a) – Solving a Quadratic Equation by Factoring 2x2 + 9x + 7 = 32x2 + 9x + 4 = 0(2x + 1)(x + 4) = 02x + 1 = x =x + 4 = x = –4The solutions are x = and x = –4. Check these in the original equation.Original equationWrite in general form.Factor.Set 1st factor equal to 0.Set 2nd factor equal to 0.
5 Example 1(b) – Solving a Quadratic Equation by Factoring cont’d6x2 – 3x = 03x(2x – 1) = 03x = x = 02x – 1 = x =The solutions are x = 0 and x = . Check these in the original equation.Original equationFactor.Set 1st factor equal to 0.Set 2nd factor equal to 0.
7 Example 2 – Extracting Square Roots Solve each equation by extracting square roots.a. 4x2 = b. (x – 3)2 = 7Solution:a. 4x2 = 12x2 = 3x = 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.
8 Example 2 – Solutioncont’dSo, the solutions are x = and x = – Check these in the original equation.b. (x – 3)2 = 7x – 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.
10 Example 3 – Completing the Square: Leading Coefficient Is 1 Solve x2 + 2x – 6 = 0 by completing the square.Solution:x2 + 2x – 6 = 0x2 + 2x = 6x2 + 2x + 12 =Write original equation.Add 6 to each side.Add 12 to each side.
11 Example 3 – Solution (x + 1)2 = 7 x + 1 = x = –1 cont’d(x + 1)2 = 7x + 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.
12 Example 3 – Solution Check: x2 + 2x – 6 = 0 cont’dCheck:x2 + 2x – 6 = 0(– )2 + 2 (– ) – 6 ≟ 08 – – – 6 ≟ 08 – 2 – 6 = 0Check the second solution in the original equation.Write original equation.Substitute – for x.Multiply.Solution checks.
14 The Quadratic FormulaThe 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.”
15 The Quadratic FormulaIn 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.
16 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 = = 45, which is positive. So, the equation has two real solutions.You can solve the equation as follows.x2 + 3x – 9 = 0Write in general form.Quadratic Formula
17 Example 6 – Solution The two solutions are: cont’dThe two solutions are:Check these in the original equation.Substitute a = 1, b = 3,and c = –9.Simplify.Simplify.
19 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
20 Example 7 – Solution Verbal Model: Labels: Width of room = w (feet) Length of room = w (feet)Area of room = (square feet)Equation: w(w + 3) = 154w2 + 3w – 154 = 0(w – 11)(w + 14) = 0
21 Example 7 – Solution w – 11 = 0 w = 11 w + 14 = 0 w = –14 cont’dw – 11 = w = 11w + 14 = w = –14Choosing the positive value, you find that the width is 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.
22 ApplicationsAnother 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 forms = –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).
23 ApplicationsA 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.
24 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