Transcript Slide 1
Warm Up Find each square root. 1. 6 2. –25 3. Solve each equation. 5. –6x = –60 x = 10 7. 2x – 40 = 0 x = 20 11 4. 6. x = 80 8. 5x = 3 California Standards 2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules for exponents. 23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. Some quadratic equations cannot be easily solved by factoring. Square roots can be used to solve some of these quadratic equations. Recall from Lesson 1-5 that every positive real number has two square roots, one positive and one negative. (Remember also that the symbol indicates a nonnegative square root.) Positive square root of 9 Negative square root of 9 When you take the square root to solve an equation, you must find both the positive and negative square root. This is indicated by the symbol ±√ . Positive and negative square roots of 9 Reading Math The expression ±3 means “3 or –3” and is read “plus or minus three.” Solve using square roots. x2 = 169 x = ± 13 The solutions are 13 and –13. Check x2 = 169 (13)2 169 169 169 Solve for x by taking the square root of both sides. Use ± to show both square roots. Substitute 13 into the original equation. Solve using square roots. x2 = –49 There is no real solution. The solution set is the empty set, ø. There is no real number whose square is negative. Solve using square roots 1. x2 = 121 2. x2 = -25 3. x2 = 0 x = ±11 no real solution. ø x=0 4. x2 = -16 no real solution. ø 5. x2 = 100 x = ±10 6. x2 = 64 x = ±8 If a quadratic equation is not written in the form x2 = a, use inverse operations to isolate x2 before taking the square root of both sides. Solve using square roots. x2 + 7 = 32 x2 + 7 = 32 –7 –7 x2 = 25 Subtract 7 from both sides. Take the square root of both sides. Solve by using square roots. 16x2 – 49 = 0 16x2 – 49 = 0 +49 +49 Add 49 to both sides. Divide by 16 on both sides. Take the square root of both sides. Use ± to show both square roots. Solve by using square roots. 36x2 = 1 Divide by 36 on both sides. Take the square root of both sides. Use ± to show both square roots. Solve by using square roots. 100x2 + 49 = 0 100x2 + 49 = 0 –49 –49 100x2 =–49 Subtract 49 from both sides. Divide by 100 on both sides. There is no real number whose square is negative. ø Solve. Round to the nearest hundredth. 0 = 90 – x2 0 = 90 – x2 + x2 + x2 x2 = 90 Add x2 to both sides. Take the square root of both sides. Estimate The exact solutions are and . The approximate solutions are 9.49 and –9.49. Solve using square roots 1. x2 -196 = 0 x = ±14 2. 0 = 3x2 -48 x = ±4 3. 24x2 +96 = 0 no real solution. ø 4. 10x2 - 75= 15 x = ±3 5. 0 = 4x2 +144 no real solution. ø 6. 5x2 – 105 = 20 x = ±5 Application Ms. Pirzada is building a retaining wall along one of the long sides of her rectangular garden. The garden is twice as long as it is wide. It also has an area of 578 square feet. What will be the length of the retaining wall? Let x represent the width of the garden. lw = A l = 2w 2x ●x = 578 2x2 = 578 Use the formula for area of a rectangle. Length is twice the width. Substitute x for w, 2x for l, and 578 for A. Continued 2x2 = 578 Divide both sides by 2. Take the square root of both sides. x = ±17 Negative numbers are not reasonable for width, so x = 17 is the only solution that makes sense. Therefore, the length is 2w or 34 feet. Lesson Quiz: Part I Solve using square roots. Check your answers. 1. x2 – 195 = 1 ± 14 2. 4x2 – 18 = –9 3. 2x2 – 10 = –12 ø 4. Solve 0 = –5x2 + 225. Round to the nearest hundredth. ± 6.71 Lesson Quiz: Part II 5. A community swimming pool is in the shape of a trapezoid. The height of the trapezoid is twice as long as the shorter base and the longer base is twice as long as the height. The area of the pool is 3675 square feet. What is the length of the longer base? Round to the nearest foot. (Hint: Use 108 feet )