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Quadratic Equations

Quadratic Equations. ALGEBRA 2 LESSON 5-5. (For help, go to Lessons 5-2 and 5-4.). Factor each expression. 1. x 2 + 5 x – 14 2. 4 x 2 – 12x 3. 9 x 2 – 16. 5-5. Quadratic Equations. ALGEBRA 2 LESSON 5-5. (For help, go to Lessons 8-3 and 7-4.).

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Quadratic Equations

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  1. Quadratic Equations ALGEBRA 2 LESSON 5-5 (For help, go to Lessons 5-2 and 5-4.) Factor each expression. 1.x2 + 5x – 14 2. 4x2 – 12x3. 9x2 – 16 5-5

  2. Quadratic Equations ALGEBRA 2 LESSON 5-5 (For help, go to Lessons 8-3 and 7-4.) 1. Factors of –14 with a sum of 5: 7 and –2 x2 + 5x – 14 = (x + 7)(x – 2) 2. 4x2 – 12x = 4x(x – 3) 3. 9x2 – 16 = (3x)2 – (4)2 = (3x – 4)(3x + 4) 4.y = x2 – 2x – 5 axis of symmetry: x = – = 1 vertex: (1, –6) Solutions (–2) 2(1) 5-5

  3. Quadratic Equations ALGEBRA 2 LESSON 5-5 (For help, go to Lessons 8-3 and 7-4.) Solutions (continued) 5.y = x2 – 4x + 4 axis of symmetry: x = – = 2 vertex: (2, 0) 6.y = x2 – 4x axis of symmetry: x = 2 vertex: (2, –4) (–4) 2(1) 5-5

  4. Quadratic Equations Standard form of a quadratic function The standard form of a quadratic function is f(x) = ax2 + bx + c, where a ≠ 0 Zero Product Property If ab = 0, then a = 0 or b = 0. Example     If (x + 3)(x − 7) = 0, then (x + 3) = 0 or (x − 7) = 0.

  5. 3x2 – 20x – 7 = 0 Write in standard form. 3x2 – 21x + x – 7 = 0 Rewrite the bx term. 3x + 1 = 0 or x – 7 = 0 Use the Zero-Product Property. 1 3 x = – or x = 7 Solve for x. 1 3 The solutions are – and 7. Quadratic Equations ALGEBRA 2 LESSON 5-5 3x2 – 20x – 7 = 0 3x(x – 7) + (x – 7) = 0 Find common factors. (3x + 1)(x – 7) = 0 Factor using the Distributive Property. 5-5

  6. Quadratic Equations ALGEBRA 2 LESSON 5-5 (continued) Check: 3x2 – 20x = 7 3x2 – 20x = 7 1 3 1 3 3 –2 – 20 – 7 3(7)2 – 20(7) 7 1 3 20 3 + 7 147 – 140 7 7 = 7 7 = 7 5-5

  7. 6x2 6 486 6 = Isolate x2. Quadratic Equations ALGEBRA 2 LESSON 5-5 Solve 6x2 – 486 = 0. 6x2 – 486 = 0 Rewrite in the form ax2 = c. x2 = 81 Simplify. x= ±9 Take the square root of each side. 5-5

  8. –270 = –16x2Isolate x2. 16.875 = x2 ±4.1 xTake the square root of each side. Quadratic Equations ALGEBRA 2 LESSON 5-5 The function y = –16x2 + 270 models the height y in feet of a heavy object x seconds after it is dropped from the top of a building that is 270 feet tall. How long does it take the object to hit the ground? y = –16x2 + 270 0 = –16x2 + 270 Substitute 0 for y. The object takes about 4.1 seconds to hit the ground. Check: Is the answer reasonable? The negative number –4.1 is also a solution to the equation. However, since a negative value for time has no meaning in this case, only the positive solution is reasonable. 5-5

  9. Define: Let x = the shorter leg. Then x + 1 = the longer leg. Write: 62 = x2 + ( x + 1 )2 36 = x2 + x2 + 2x + 1 Multiply. 0 = 2x2 + 2x – 35 Write in standard form. Quadratic Equations ALGEBRA 2 LESSON 5-5 A carpenter wants to cut a piece of plywood in the shape of a right triangle. He wants the hypotenuse of the triangle to be 6 feet long, as shown in the diagram. About how long should the perpendicular sides be? Relate: From the Pythagorean Theorem we know for a right triangle that the hypotenuse squared equals the sum of the squares of the other two sides. 5-5

  10. Quadratic Equations ALGEBRA 2 LESSON 5-5 (continued) Graph the related function y = 2x2 + 2x – 35. Use the CALC feature to find the positive solution. The sides of the triangle are 3.7 ft and 4.7 ft. 5-5

  11. Assignment Page 266 2-18 even, 32, 35-47 odd

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