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1.4 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

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1.4 - 21.1 - 2 Quadratic Equations 1.4 Solving a Quadratic Equation Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant

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1.4 - 3 Quadratic Equation in One Variable An equation that can be written in the form where a, b, and c are real numbers with a 0, is a quadratic equation. The given form is called standard form.

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1.4 - 4 Second-degree Equation A quadratic equation is a second-degree equation. This is an equation with a squared variable term and no terms of greater degree.

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1.4 - 5 Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both.

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1.4 - 6 Example 1 USING THE ZERO-FACTOR PROPERTY Solve Solution: Standard form Factor. Zero-factor property.

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1.4 - 7 Example 1 USING THE ZERO-FACTOR PROPERTY Solve Solution: Zero-factor property. Solve each equation.

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1.4 - 8 Square-Root Property A quadratic equation of the form x 2 = k can also be solved by factoring Subtract k. Factor. Zero-factor property. Solve each equation.

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1.4 - 9 Square Root Property If x 2 = k, then

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1.4 - 10 Square-Root Property That is, the solution of If k = 0, then this is sometimes called a double solution. Both solutions are real if k > 0, and both are imaginary if k < 0 If k < 0, we write the solution set as

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1.4 - 11 Example 2 USING THE SQUARE ROOT PROPERTY a. Solution: By the square root property, the solution set is Solve each quadratic equation.

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1.4 - 12 Example 2 USING THE SQUARE ROOT PROPERTY b. Solution: Since the solution set of x 2 = 25 is Solve each quadratic equation.

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1.4 - 13 Example 2 USING THE SQUARE ROOT PROPERTY c. Solution: Use a generalization of the square root property. Generalized square root property. Add 4. Solve each quadratic equation.

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1.4 - 14 Solving A Quadratic Equation By Completing The Square To solve ax 2 + bx + c = 0, by completing the square: Step 1 If a 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Square half the coefficient of x, and add this square to both sides of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution.

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1.4 - 15 Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x 2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Step 2 Add 14 to both sides. Step 3 add 4 to both sides. Step 4 Factor; combine terms.

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1.4 - 16 Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x 2 – 4x –14 = 0 by completing the square. Solution Step 4 Factor; combine terms. Step 5 Square root property. Take both roots. Add 2. Simplify the radical. The solution set is

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1.4 - 17 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Divide by 9. (Step 1) Add – 1. (Step 2)

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1.4 - 18 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Factor, combine terms. (Step 4) Square root property

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1.4 - 19 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x 2 – 12x + 9 = 0 by completing the square. Solution Quotient rule for radicals Square root property Add.

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1.4 - 20 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 The solution set is Solution Add. Solve 9x 2 – 12x + 9 = 0 by completing the square.

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1.4 - 21 The Quadratic Formula The method of completing the square can be used to solve any quadratic equation. If we start with the general quadratic equation, ax 2 + bx + c = 0, a 0, and complete the square to solve this equation for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. We assume that a > 0.

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1.4 - 22 Quadratic Formula The solutions of the quadratic equation ax 2 + bx + c = 0, where a 0, are

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1.4 - 23 Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

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1.4 - 24 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Write in standard form. Here a = 1, b = – 4, c = 2 Quadratic formula.

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1.4 - 25 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Quadratic formula. The fraction bar extends under – b.

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1.4 - 26 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: The fraction bar extends under – b.

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1.4 - 27 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x 2 – 4x = – 2 Solution: Factor first, then divide. Factor out 2 in the numerator. Lowest terms. The solution set is

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1.4 - 28 Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x 2 = x – 4. Solution: Write in standard form. Quadratic formula; a = 2, b = – 1, c = 4 Use parentheses and substitute carefully to avoid errors.

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1.4 - 29 Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x 2 = x – 4. Solution: The solution set is

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1.4 - 30 Cubic Equation The equation x 3 + 8 = 0 that follows is called a cubic equation because of the degree 3 term. Some higher-degree equations can be solved using factoring and the quadratic formula.

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1.4 - 31 Example 7 SOLVING A CUBIC EQUATION Solve Solution Factor as a sum of cubes. or Zero-factor property or Quadratic formula; a = 1, b = – 2, c = 4

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1.4 - 32 Example 7 SOLVING A CUBIC EQUATION Solve Solution Simplify. Simplify the radical. Factor out 2 in the numerator.

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1.4 - 33 Example 7 SOLVING A CUBIC EQUATION Solve Solution Lowest terms

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1.4 - 34 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve for the specified variable. Use when taking square roots. Solution a. Goal: Isolate d, the specified variable. Multiply by 4. Divide by.

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1.4 - 35 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. See the Note following this example. Square root property Divide by.

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1.4 - 36 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Square root property Rationalize the denominator.

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1.4 - 37 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Rationalize the denominator. Multiply numerators; multiply denominators.

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1.4 - 38 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution a. Multiply numerators; multiply denominators. Simplify.

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1.4 - 39 Solving for a Specified Variable Note In Example 8, we took both positive and negative square roots. However, if the variable represents a distance or length in an application, we would consider only the positive square root.

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1.4 - 40 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. Write in standard form. Now use the quadratic formula to find t.

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1.4 - 41 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. a = r, b = – s, and c = – k

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1.4 - 42 Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Solve the specified variable. Use when taking square roots. Solution b. a = r, b = – s, and c = – k Simplify.

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1.4 - 43 The Discriminant The Discriminant The quantity under the radical in the quadratic formula, b 2 – 4ac, is called the discriminant. Discriminant

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1.4 - 44 The Discriminant Discriminant Number of Solutions Type of Solution Positive, perfect square TwoRational Positive, but not a perfect square TwoIrrational Zero One (a double solution) Rational NegativeTwo Nonreal complex

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1.4 - 45 Caution The restriction on a, b, and c is important. For example, for the equation the discriminant is b 2 – 4ac = 5 + 4 = 9, which would indicate two rational solutions if the coefficients were integers. By the quadratic formula, however, the two solutions are irrational numbers,

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1.4 - 46 Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution a. For 5x 2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is b 2 – 4ac = 2 2 – 4(5)(– 4) = 84 The discriminant is positive and not a perfect square, so there are two distinct irrational solutions.

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1.4 - 47 Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution b. First write the equation in standard form as x 2 – 10x + 25 = 0. Thus, a = 1, b = – 10, and c = 25, and b 2 – 4ac =(– 10 ) 2 – 4(1)(25) = 0 There is one distinct rational solution, a double solution.

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1.4 - 48 Example 9 USING THE DISCRIMINANT Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. Solution c. For 2x 2 – x + 1 = 0, a = 2, b = –1, and c = 1, so b 2 – 4ac = (–1) 2 – 4(2)(1) = –7. There are two distinct nonreal complex solutions. (They are complex conjugates.)

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