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Lesson Menu Five-Minute Check (over Lesson 4–5) CCSS Then/Now New Vocabulary Key Concept: Quadratic Formula Example 1:Two Rational Roots Example 2:One Rational Root Example 3:Irrational Roots Example 4:Complex Roots Key Concept: Discriminant Example 5:Describe Roots Concept Summary: Solving Quadratic Equations

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Over Lesson 4–5 5-Minute Check 1 A.4, 2 B.4, –2 C.2, –2 D.2, 1 Solve x 2 – 2x + 1 = 9 by using the Square Root Property.

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Over Lesson 4–5 5-Minute Check 2 Solve 4c 2 + 12c + 9 = 7 by using the Square Root Property. A.–3, 3 B. C.–1, 1 D.

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Over Lesson 4–5 5-Minute Check 3 Find the value of c that makes the trinomial x 2 + x + c a perfect square. Then write the trinomial as a perfect square. A. B. C. D.1; (x + 1) 2

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Over Lesson 4–5 5-Minute Check 4 Solve x 2 + 2x + 24 = 0 by completing the square. A.–12, 12 B.–2, 2 C. D.

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Over Lesson 4–5 5-Minute Check 5 Solve 5g 2 – 8 = 6g by completing the square. A.2, –1 B. C. D.4, 2

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Over Lesson 4–5 5-Minute Check 6 A.–10, 10 B.5, 20 C.–20, 20 D.4, 25 Find the value(s) of k in x 2 + kx + 100 = 0 that would make the left side of the equation a perfect square trinomial.

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CCSS Content Standards N.CN.7 Solve quadratic equations with real coefficients that have complex solutions. A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity. Mathematical Practices 8 Look for and express regularity in repeated reasoning.

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Then/Now You solved equation by completing the square. Solve quadratic equations by using the Quadratic Formula. Use the discriminant to determine the number and type of roots of a quadratic equation.

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Vocabulary Quadratic Formula discriminant

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Concept

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Example 1 Two Rational Roots Solve x 2 – 8x = 33 by using the Quadratic Formula. First, write the equation in the form ax 2 + bx + c = 0 and identify a, b, and c. x 2 – 8x = 331x 2 – 8x – 33 = 0 ax 2 + bx + c = 0 Then, substitute these values into the Quadratic Formula. Quadratic Formula

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Example 1 Two Rational Roots Replace a with 1, b with –8, and c with –33. Simplify.

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Example 1 Two Rational Roots x = 11 x= –3Simplify. Answer: The solutions are 11 and –3. orWrite as two equations.

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Example 1 A.15, –2 B.2, –15 C.5, –6 D.–5, 6 Solve x 2 + 13x = 30 by using the Quadratic Formula.

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Example 2 One Rational Root Solve x 2 – 34x + 289 = 0 by using the Quadratic Formula. Identify a, b, and c. Then, substitute these values into the Quadratic Formula. Quadratic Formula Replace a with 1, b with –34, and c with 289. Simplify.

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Example 2 One Rational Root CheckA graph of the related function shows that there is one solution at x = 17. Answer: The solution is 17. [–5, 25] scl: 1 by [–5, 15] scl: 1

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Example 2 A.11 B.–11, 11 C.–11 D.22 Solve x 2 – 22x + 121 = 0 by using the Quadratic Formula.

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Example 3 Irrational Roots Solve x 2 – 6x + 2 = 0 by using the Quadratic Formula. Quadratic Formula Replace a with 1, b with –6, and c with 2. Simplify. or

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Example 3 Irrational Roots CheckCheck these results by graphing the related quadratic function, y = x 2 – 6x + 2. Using the ZERO function of a graphing calculator, the approximate zeros of the related function are 0.4 and 5.6. Answer: [–10, 10] scl: 1 by [–10, 10] scl: 1

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A. B. C. D. Example 3 Solve x 2 – 5x + 3 = 0 by using the Quadratic Formula.

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Example 4 Complex Roots Solve x 2 + 13 = 6x by using the Quadratic Formula. Quadratic Formula Replace a with 1, b with –6, and c with 13. Simplify.

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Example 4 Complex Roots A graph of the related function shows that the solutions are complex, but it cannot help you find them. Answer: The solutions are the complex numbers 3 + 2i and 3 – 2i. [–5, 15] scl: 1 by [–5, 15] scl: 1

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Example 4 Complex Roots x 2 + 13=6xOriginal equation Check To check complex solutions, you must substitute them into the original equation. The check for 3 + 2i is shown below. (3 + 2i) 2 + 13=6(3 + 2i)x = (3 + 2i) ? 9 + 12i + 4i 2 + 13 =18 + 12iSquare of a sum; Distributive Property ? 22 + 12i – 4=18 + 12iSimplify. ? 18 + 12i =18 + 12i

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Example 4 A.2 ± i B.–2 ± i C.2 + 2i D.–2 ± 2i Solve x 2 + 5 = 4x by using the Quadratic Formula.

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Concept

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Example 5 Describe Roots A. Find the value of the discriminant for x 2 + 3x + 5 = 0. Then describe the number and type of roots for the equation. a = 1, b = 3, c = 5 b 2 – 4ac=(3) 2 – 4(1)(5)Substitution = 9 – 20Simplify. = –11Subtract. Answer: The discriminant is negative, so there are two complex roots.

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Example 5 Describe Roots B. Find the value of the discriminant for x 2 – 11x + 10 = 0. Then describe the number and type of roots for the equation. a = 1, b = –11, c = 10 b 2 – 4ac=(–11) 2 – 4(1)(10)Substitution = 121 – 40Simplify. = 81Subtract. Answer: The discriminant is 81, so there are two rational roots.

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Example 5 A.0; 1 rational root B.16; 2 rational roots C.32; 2 irrational roots D.–64; 2 complex roots A. Find the value of the discriminant for x 2 + 8x + 16 = 0. Describe the number and type of roots for the equation.

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Example 5 A.0; 1 rational root B.36; 2 rational roots C.32; 2 irrational roots D.–24; 2 complex roots B. Find the value of the discriminant for x 2 + 2x + 7 = 0. Describe the number and type of roots for the equation.

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Concept

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End of the Lesson

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