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Solving Quadratic Equations Section 1.3 What is a Quadratic Equation? A quadratic equation in x is an equation that can be written in the standard form: ax² + bx + c = 0 Where a,b,and c are real numbers and a ≠ 0. Solving a Quadratic Equation by Factoring. The factoring method applies the zero product property: Words: If a product is zero, then at least one of its factors has to be zero. Math: If (B)(C)=0, then B=0 or C=0 or both. Recap of steps for how to solve by Factoring Set equal to 0 Factor Set each factor equal to 0 (keep the squared term positive) Solve each equation (be careful when determining solutions, some may be imaginary numbers) Example 1 Solve x² - 12x + 35 = 0 by factoring. Factor: (x – 7)(x - 5) = 0 Set each factor equal to zero by the zero product property. (x – 7)=0 Solve each equation to find solutions. x = 7 or x = 5 The solution set is: { 5, 7 } (x – 5)=0 Example 2 Solve 3t² + 10t + 6 = -2 by factoring. Check equation to make sure it is in standard form before solving. Is it? It is not, so set equation equal to zero first: 3t² + 10t + 8 = 0 Now factor and solve. (3t + 4)(t + 2) = 0 3t + 4 = 0 t= 4 3 t +2 = 0 t = -2 Solve by factoring. 2x 3x 0 x 2 x 3 0 2 x0 2x 3 0 x 3 2 Solve by the Square Root Method. If the quadratic has the form ax² + c = 0, where a ≠ 0, then we could use the square root method to solve. Words: If an expression squared is equal to a constant, then that expression is equal to the positive or negative square root of the constant. Math: If x² = c, then x = ±c. Note: The variable squared must be isolated first (coefficient equal to 1). Example 1: Solve by the Square Root Method: 2x² - 32 = 0 2x² = 32 x² = 16 2 x = 16 x=±4 Example 2: Solve by the Square Root Method. 5x² + 10 = 0 5x² = -10 x² = -2 x=± 2 x = ±i 2 Example 3: Solve by the Square Root Method. (x – 3)² = 25 x–3=±5 x – 3 = 5 or x – 3 = -5 x=8 x = -2 Solve by the Square Root Method 3 x 8 2 12 3 x 8 12 3x 8 2 3 3 x 8 2 3 3 x 8 2 3 3 x 8 2 3 82 3 82 3 x 3 x 3 Solve by Completing the Square. Words Express the quadratic equation in the following form. Divide b by2 and square the result, then add the square to both sides. Write the left side of the equation as a perfect square. Solve by using the square root method. Math x² + bx = c x² + bx + ( b 2 )² = c + ( (x + b )² = c + ( b )² 2 2 b 2 )² Example 1: Solve by Completing the Square. x² + 8x – 3 = 0 x² + 8x = 3 x² + 8x + (4)² = 3 + (4)² x² + 8x + 16 = 3 + 16 (x + 4)² = 19 x + 4 = ± 19 x = -4 ± 19 Add three to both sides. Add ( b )² which is (4)² to both sides. 2 Write the left side as a perfect square and simplify the right side. Apply the square root method to solve. Subtract 4 from both sides to get your two solutions. Example 2: Solve by Completing the Square when the Leading Coefficient is not equal to 1. 2x² - 4x + 3 = 0 x² - 2x + 3 Divide by the leading coefficient. Continue to solve using the completing the square method. Simplify radical. =0 2 x² - 2x + ___ = 3 + ____ 2 x² - 2x + 1 = 3 + 1 (x – 1)² = x–1=± x=1± 5 2 5 2 10 2 2 Quadratic Formula If a quadratic can’t be factored, you must use the quadratic formula. If ax² + bx + c = 0, then the solution is: x b b 4 ac 2 2a a=1 b = -4 Solve x x 4x 1 0 2 b b 4 ac 2 2a x x 4 4 x 2 42 5 4 1 1 2 2 1 4 16 4 2 x c = -1 4 20 2 x2 5 Solve 4n 6n 9 2 4n 6n 9 0 2 n b b 4 ac 2 n 6 180 8 2a n n 6 6 4 4 9 2 n 66 5 8 24 6 36 144 8 n 33 5 4 x 8 4x 2 Solve x 4x 8 0 2 x b x 4 2 b 4 ac 2 2a 16 x 4 4i 2 x 4 x 4 2 4 18 2 1 4 16 32 2 x 2 2i Discriminant x b b 4 ac 2 2a The term inside the radical b² - 4ac is called the discriminant. The discriminant gives important information about the corresponding solutions or answers of ax² + bx + c = 0, where a,b, and c are real numbers. b² - 4ac b² - 4ac > 0 b² - 4ac = 0 b² - 4ac < 0 Solutions Tell what kind of solution to expect x 28 x 198 0 2 b 4 ac 28 4 1198 2 2 784 792 8