University:
California State University, Northridge
Course:
MATH 360 | Abstract Algebra I
Academic year:

2024
Views:

447

Pages:

18
Author:

razkorak36n

Chapter 15 Roots and Radicals § 15.5 Solving Equations Containing Radicals Extraneous Solutions Power Rule (text only talks about squaring, but applies to other powers, as well). If both sides of an equation are raised to the same power, solutions of the new equation contain all the solutions of the original equation, but might also contain additional solutions. A proposed solution of the new equation that is NOT a solution of the original equation is an extraneous solution. 35 Solving Radical Equations Example Solve the following radical equation. ( x +1 = 5 ) 2 x + 1 = 52 x + 1 = 25 x = 24 Substitute into the original equation. 24 + 1 = 5 25 = 5 true So the solution is x = 24. 36 Solving Radical Equations Example Solve the following radical equation. Substitute into the 5 x = −5 original equation. ( 5x ) = (− 5) 2 5 x = 25 2 5 ⋅ 5 =−5 25 = −5 Does NOT check, since the left side of the equation is asking for the x=5 principal square root. So the solution is ∅. 37 Solving Radical Equations Steps for Solving Radical Equations 1) Isolate one radical on one side of equal sign. 2) Raise each side of the equation to a power equal to the index of the isolated radical, and simplify. (With square roots, the index is 2, so square both sides.) 3) If equation still contains a radical, repeat steps 1 and 2. If not, solve equation. 4) Check proposed solutions in the original equation. 38 Solving Radical Equations Example Solve the following radical equation. x −1 −1 = 0 ( x −1 = 1 ) Substitute into the original equation. x − 1 = 12 2 − 1 − 1 =0 x −1 = 1 1 −1 = 0 2 x=2 1 − 1 = 0 true So the solution is x = 2. 39 Solving Radical Equations Example Solve the following radical equation. 2x + x +1 = 8 ( x +1 = 8 − 2x ) 2 x + 1 = (8 − 2 x ) 2 x + 1 = 64 − 32 x + 4 x 2 2 0 = 63 − 33 x + 4 x 0 = (3 − x)(21 − 4 x) 21 x = 3 or 4 40 Solving Radical Equations Example continued Substitute the value for x into the original equation, to check the solution. 2(3) + 3 + 1 = 8 6 + 4 = 8 true So the solution is x = 3. ( ) 21 21 2 + +1 = 8 4 4 21 25 + =8 2 4 21 5 + =8 2 2 26 =8 2 false 41 Solving Radical Equations Example Solve the following radical equation. y+5 = 2− y−4 ( ) ( 2 y+5 = 2− y−4 ) 2 y+5 = 4−4 y−4 + y−4 5 = −4 y − 4 5 − = 4 2  5 −  =  4 y−4 ( y−4 25 = y−4 16 25 89 y = 4+ = 16 16 ) 2 42 Solving Radical Equations Example continued Substitute the value for x into the original equation, to check the solution. 89 89 +5 = 2− −4 16 16 169 25 = 2− 16 16 13 5 = 2− 4 4 13 3 = 4 4 false So the solution is ∅. 43 Solving Radical Equations Example Solve the following radical equation. 2 x − 4 − 3 x + 4 = −2 ( 2 x − 4 = −2 + 3 x + 4 ) ( 2 2 x − 4 = − 2 + 3x + 4 ) 2 2 x − 4 = 4 − 4 3x + 4 + 3x + 4 2 x − 4 − 8 − 3 x = −4 3 x + 4 x 2 − 24 x + 80 = 0 − x − 12 = −4 3 x + 4 (x − 20)(x − 4) = 0 (− x − 12) 2 ( ) 2 = − 4 3x + 4 x 2 + 24 x + 144 = 16(3 x + 4) = 48 x + 64 x = 4 or 20 44 Solving Radical Equations Example continued Substitute the value for x into the original equation, to check the solution. 2(4) − 4 − 3(4) + 4 =−2 2(20) − 4 − 3(20) + 4 =−2 4 − 16 = −2 36 − 64 = −2 2 − 4 = −2 6 − 8 = −2 true true So the solution is x = 4 or 20. 45 § 15.6 Radical Equations and Problem Solving The Pythagorean Theorem Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. (leg a)2 + (leg b)2 = (hypotenuse)2 leg a hypotenuse leg b 47 Using the Pythagorean Theorem Example Find the length of the hypotenuse of a right triangle when the length of the two legs are 2 inches and 7 inches. c2 = 22 + 72 = 4 + 49 = 53 c= 53 inches 48 The Distance Formula By using the Pythagorean Theorem, we can derive a formula for finding the distance between two points with coordinates (x1,y1) and (x2,y2). d= (x2 − x1 ) + ( y2 − y1 ) 2 2 49 The Distance Formula Example Find the distance between (−5, 8) and (−2, 2). d= (x2 − x1 ) + ( y2 − y1 ) d= (− 5 − (−2) ) + (8 − 2) d= (− 3) + (6) 2 2 2 2 2 2 d = 9 + 36 = 45 = 3 5 50