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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Warm Up (Add to HW & Pass Back Papers) Evaluate for x =–2, y = 3, and z = –1. 6 1. x 2 2. xyz 3. x 2 – yz4. y – xz 4 71

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant 9-9 The Quadratic Formula and the Discriminant Holt Algebra 1

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9-9 The Quadratic Formula and the Discriminant

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Example 1A: Using the Quadratic Formula Solve using the Quadratic Formula. 6x 2 + 5x – 4 = 0 6x 2 + 5x + (–4) = 0 Identify a, b, and c. Use the Quadratic Formula. Simplify. Substitute 6 for a, 5 for b, and –4 for c.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Example 1A Continued Solve using the Quadratic Formula. 6x 2 + 5x – 4 = 0 Simplify. Write as two equations. Solve each equation.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Example 1B: Using the Quadratic Formula Solve using the Quadratic Formula. x 2 = x + 20 1x 2 + (–1x) + (–20) = 0 Write in standard form. Identify a, b, and c. Use the quadratic formula. Simplify. Substitute 1 for a, –1 for b, and –20 for c.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Example 1B Continued Solve using the Quadratic Formula. x = 5 or x = –4 Simplify. Write as two equations. Solve each equation. x 2 = x + 20

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Many quadratic equations can be solved by graphing, factoring, taking the square root, or completing the square. Some cannot be solved by any of these methods, but you can always use the Quadratic Formula to solve any quadratic equation.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant If the quadratic equation is in standard form, the discriminant of a quadratic equation is b 2 – 4ac, the part of the equation under the radical sign.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant 3x 2 – 2x + 2 = 0 2x 2 + 11x + 12 = 0 x 2 + 8x + 16 = 0 a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16 b 2 – 4ac (–2) 2 – 4(3)(2) 11 2 – 4(2)(12)8 2 – 4(1)(16) 4 – 24121 – 9664 – 64 –2025 0 b 2 – 4ac is negative. There are no real solutions b 2 – 4ac is positive. There are two real solutions b 2 – 4ac is zero. There is one real solution Example 3: Using the Discriminant Find the number of solutions of each equation using the discriminant. A.B.C.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Example 5: Solving Using Different Methods Solve x 2 – 9x + 20 = 0. Show your work. Method 1 Solve by graphing. y = x 2 – 9x + 20 Write the related quadratic function and graph it. The solutions are the x-intercepts, 4 and 5.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Solve x 2 – 9x + 20 = 0. Show your work. Method 2 Solve by factoring. x 2 – 9x + 20 = 0 Example 5 Continued (x – 5)(x – 4) = 0 x – 5 = 0 or x – 4 = 0 x = 5 or x = 4 Factor. Use the Zero Product Property. Solve each equation.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Solve x 2 – 9x + 20 = 0. Show your work. Method 3 Solve by completing the square. Example 5 Continued x 2 – 9x + 20 = 0 Add to both sides. x 2 – 9x = –20 x 2 – 9x + = –20 + Factor and simplify. Take the square root of both sides.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Solve x 2 – 9x + 20 = 0. Show your work. Method 3 Solve by completing the square. Example 5 Continued Solve each equation. x = 5 or x = 4

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Method 4 Solve using the Quadratic Formula. Example 5: Solving Using Different Methods. 1x 2 – 9x + 20 = 0 x = 5 or x = 4 Identify a, b, c. Substitute 1 for a, –9 for b, and 20 for c. Simplify. Write as two equations. Solve each equation. Solve x 2 – 9x + 20 = 0. Show your work.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Lesson Quiz: Part I 1. Solve –3x 2 + 5x = 1 by using the Quadratic Formula. 2. Find the number of solutions of 5x 2 – 10x – 8 = 0 by using the discriminant. ≈ 0.23, ≈ 1.43 2

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Lesson Quiz: Part II The discriminant is zero. The object will reach its maximum height of 40 feet once. 4. Solve 8x 2 – 13x – 6 = 0. Show your work. 3. The height h in feet of an object shot straight up is modeled by h = –16t 2 + vt + c, where c is the beginning height of the object above the ground. An object is shot up from 4 feet off the ground with an initial velocity of 48 feet per second. Will it reach a height of 40 feet? Use the discriminant to explain your answer.

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Holt Algebra 1 9-9 The Quadratic Formula and the Discriminant Warm-Up 1. Solve –3x 2 + 5x = 1 by using the Quadratic Formula. 2. Find the number of solutions of 5x 2 – 10x – 8 = 0 by using the discriminant. ≈ 0.23, ≈ 1.43 2

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