CH 60 MORE PYTHAGOREAN THEOREM AND MORE RADICALS Introduction T he last time we studied the Pythagorean Theorem we may have used our calculator to round square roots that didn t come out whole numbers. But rounding off square roots is not always a good idea in the real world. After we ve done some applied mathematics, we may give the results to an engineer. Suppose we ve rounded to the hundredths place (good enough for business) but the engineer is designing the Core 9 Quad microprocessor, and she needs all the digits of the numbers out to the billionths place. We haven t done her any good at all by rounding off for her. The safest thing to do is to give her the exact results and let her round according to her requirements. We re now going to work with the Pythagorean Theorem again, but this time we re going to leave our answers exact, using the methods of the previous chapter. To this end, we need to review the key concept regarding square roots. Important Fact About Square Roots If we square the square root of something (that s not negative), we get the something : ( ) ( 4) (.809189) 4
6 And we can write the general formula: x x Squaring Undoes Square-Rooting Another skill we ll need for this chapter is the ability to simplify an expression like exponents: (ab) a b. ( ). Recall one of our five laws of This rule means that we can simplify ( ) ( ) ( ) as follows: Homework 1. While it s the case that (ab) a b, is this rule true for addition? That is, does (a + b) a + b?. Simplify each expression: a. ( ) b. ( 9) c. ( ) d. ( 101) e. ( ) f. ( ) g. (9 ) h. (1 )
Solving Right Triangles Recall the Pythagorean Theorem: a 90 b c a b c EXAMPLE 1: Solve each right triangle problem: A. The legs are 6 and 10. Find the hypotenuse. a + b c 6 + 10 c 6 + 100 c c 16 c 16 4 4 4 Notice that c 16 is a quadratic equation, and therefore probably has two solutions; in fact, the solutions are 16, or 4. But in this problem we re talking about the hypotenuse of a triangle, whose length must be positive. So we immediately discard the negative solution and retain just the positive one. B. One leg is 0 and the hypotenuse is 0. Find the other leg. a + b c 0 + b 0 400 + b 900 b 00 b 00 100 10
8 C. The legs are and 1. Find the hypotenuse. a + b c 1 c ( ) ( ) 8 + 11 c c 14 c 14 D. The hypotenuse is and one of the legs is 8. Find the other leg. a + b c a 8 ( ) a + 64 a 8 a 8 4 E. One of the legs is and the hypotenuse is 4. Find the other leg. a + b c a + ( ) ( 4 ) a + 18 4 a 4 a 4 6 Homework Calculate the EXACT answer to each Pythagorean problem:. The legs of a right triangle are 8 and 1. Find the hypotenuse. 4. The legs of a right triangle are 6 and 9. Find the hypotenuse.. Find the hypotenuse of a right triangle given that its legs have lengths of and. 6. One leg of a right triangle is 1 and the hypotenuse is 0. Find the other leg.. The hypotenuse of a right triangle is 18 and one of its legs is 10. Find the other leg.
9 8. One leg of a right triangle is and the hypotenuse is 4. Find the other leg. 9. Find the missing leg of a right triangle if the hypotenuse is 18 and one of its legs is. 10. The legs of a right triangle are and. Find the hypotenuse. 11. The hypotenuse of a right triangle is and one of its legs is. Find the length of the other leg. 1. A leg of a right triangle is 8 and its hypotenuse is 10. Find the length of the other leg. 1. The leg of a right triangle is and its hypotenuse is 4. Find the other leg. Review of Simplifying Radicals Recall how we simplify 6 : 6 9 9 If we come across 11, we simply change it to 11, and we know that 0 0. And when asked to simplify 1, we don t even try because this does not exist in Elementary Algebra. Also remember that means plus or minus. Thus, to evaluate the expression 10, we write 10 + 1 (using the plus sign) 10 8 (using the minus sign) So 10 represents the two numbers 1 and 8. Just as 8, if we see the expression 1 write it as. 0, we should probably
60 Last example in this section: To simplify 4 81, and therefore, 10 4, we notice that 10 4 10 81 10 9 This final answer can also be written separately as 10 9, 10 9 The next section of this chapter will deal with the possibility that after we have simplified the radical, we may need to continue simplifying by reducing the fraction to lowest terms. Homework 14. Simplify each expression: a. 1 b. 100 144 c. 0 d. 9 e. 0 100 f. 80 g. 1 1 h. 8 144 i. 1 j. 8 k. 1 18 l. 0 m. 1 n. 4 9 o. 4 9 p. 9 q. 14 r. s. 0 t. 10 0 u. 4 8 v. 4 8 w. 1 196 x. 1 9 y. 0 z. 8 6
61 Reducing Fractions with Radicals We recall that reducing a fraction to lowest terms involves two main steps: For example, 1) Factor the top and bottom ) Divide out (cancel) any common factors terms 1 term 8 6 x (4 x) (4 x) 4 () () 4 x EXAMPLE : Simplify: 1 0 Solution: expression: We have all the skills required to simplify this 1 0 1 0 1 0 1 0 ( ) ( ) (the original fraction) (factor using a perfect square) (split the radical in two) (simplify the radical) (factor the top and bottom) (divide out the common factor of ) A little arithmetic gives the final reduced fraction 4
6 EXAMPLE : Simplify each radical expression: A. 14 (49)() B. 00 100 10 1 1 1 Homework 1. Simplify each expression: a. 1 100 4 b. c. 40 6 d. 4 8 e. 18 18 6 f. 00 10 g. 8 6 h. 0 0 1 i. 98 j. 0 10 k. 0 144 6 l. 0 00 0 m. 90 14 n. 8 0 4 o. 8 1 8 p. 9 8 8 q. 8 r. 4
6 Review Problems 16. In each problem, two of the sides of a right triangle are given (l leg and h hypotenuse). Find the third side. a. l ; l 10 b. l 1; l c. l 4; l 6 d. l 8; l e. l 4; l 10 f. l ; l 14 g. l ; h 1 h. l ; h i. l 6; h j. l ; h 1 k. l ; h l. l ; h 4 m. l ; l n. l ; h 8 o. l ; h 1 p. l ; h 6 q. l 10; h 41 r. l 11; h 8 s. l 1; l 1 t. l ; l 6 u. l 4; l v. l 1; h w. l 1; l 1 x. l ; h y. l ; h 1 z. l ; h 0
64 Simplify each expression: 1. 10 0 1 18. 19. 14 14 8 0. 1 9 4 1. 10 14. 8. 14 14 14 4. 98 1 Solutions 1. NO! You should know quite well that (a + b) (a + b)(a + b) a + ab + ba + b a + ab + b. a. b. 9 c. d. 101 e. 1 f. 1 g. 16 h. 1008. 4 1 4. 1. 4 6.. 4 14 8. 9. 10. 0 11. 16 1. 1 1. This scenario is impossible, since the hypotenuse of a right triangle must be longer than either of its legs. Here s a mathematical proof: a + b c a + 4 (presuming that the leg is and the hypotenuse is 4) a + 16
6 a 9 (subtract from each side of the equation) a 9, which is not a number in this class. And even if it becomes a number in a later class, it still couldn t be considered the length of a side of a triangle. 14. a. 1, b. 44, 44 c. d., 1 e. 10 f. 4 g., 0 h. 4, 0 i. 1 j. k. 1 l. m. q. u. y. 1 4 n. Does not exist o. r. v. z. 1 s. w. 8 8 11 4 9 p. 9 10 t., 1 x. Does not exist 1. a. 11 16 10 b. c. 6 1 d. e. f. 1 4 g. h. i. 1 j. k. l. 10 m. Does not exist n. 1 o. Does not exist p. q. r. 4 16. a. 149 b. 10 c. 1 d. 4 e. 6 f. 4 g. 119 h. i. 14 j. 6 k. l. m. n. 1 o. p. 1 q. 1 r. s. 6 t. 1
66 u. v. w. x. y. 4 z. 8 1. 0. Does not exist 1.. 14 14 14 18. 19. 4. 10 14. 4 He who opens a school door, closes a prison. Victor Hugo