Ch5_Radicals - MA100 Fall 2010 PART 1 Precalculus[5...

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Unformatted text preview: MA100, Fall 2010, PART 1: Precalculus [5] Exponents and radicals 1. Exponents and their rules Let a > 0 and n ∈ N . The number a n = a · a · a . . . · a ( n times) denotes the n th power of a . The number a is called the base , and n is called the exponent . By definition, we have following: a = 1 and a- n = 1 a n (1) Let a, b > 0 and r, s ∈ R . We have the following rules: a r a s = a r + s and a r a s = a r- s (2) ( a r ) s = a r s (3) ( a b ) r = a r b r and a b r = a r b r . (4) Note n = 0 and 1 n = 1 . Observation ( a + b ) r 6 = a r + b r !!! For instance, (2 + 7) 2 6 = 2 2 + 7 2 and, in general, ( a + b ) 2 6 = a 2 + b 2 ! This is a common mistake made by students. On a warning note, a mistake like this could lead to you failing this course! Note There is no such thing as 0 ! (at least for now). Example: Simplify ( x- 5 y 3 z 10 )- 3 5 . Solution: ( x- 5 y 3 z 10 )- 3 5 = ( x- 5 )- 3 5 ( y 3 )- 3 5 ( z 10 )- 3 5 by relation (4) = x- 5 × (- 3 5 ) y 3 × (- 3 5 ) z 10 × (- 3 5 ) by relation (3) = x (- 5 1 ) × (- 3 5 ) y ( 3 1 ) × (- 3 5 ) z ( 10 1 ) × (- 3 5 ) = x 3 y- 9 5 z- 6 . 1 2. Square root radicals Definition Let a > . The positive solution of the equation x 2 = a is called the positive square root of a and it is denoted by √ a. In other words, √ a is a positive number and it fulfills ( √ a ) 2 = a . Example: √ 5 is a positive number and it fulfills √ 5 2 = 5 . Example: √ 49 is a positive number and it fulfills √ 49 2 = 49 . In this case, we have that √ 49 = 7 . Note If a > 0, the equation x 2 = a has two real solutions. When we write √ a we always refer to the positive root of this equation. Example: √ 49 = 7, and not to (- 7) (!), since 7 is the positive root of x 2 = 49. Note If a > 0, the equation x 2 = a has two real solutions. These are x = ± √ a. Example: The equation x 2 = 49 has two solutions: one positive, x = 7, and one negative x =- 7 . Since √ 49 = 7, the solutions of the equation x 2 = 49 may be written as x = ± √ 49 (or, in this case, x = ± 7) . Example: Solve x 2 = 9. Answer: x = ± √ 9 , or x = ± 3 . Example: Solve x 2 = 11. Answer: x = ± √ 11 . Example: Solve x 2 = 24. Answer: x = ± √ 24 . Note In this course we deal with real numbers only! For us, other kind of numbers do not exist . Observation Square roots of negative numbers do not exist as real numbers. This is because the equation x 2 = a has no real solutions when a < 0. For example, √- 4 is not a real number since the equation x 2 =- 4 has no real solution. (For all numbers x we have x 2 ≥ 0, whereas- 4 < . ) So, remember: Square roots of negative numbers do not exist ! Example: Solve x 2 =- 3 . Answer: “this equation has no real number solutions”, or, simply, ”there are no solutions”....
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This note was uploaded on 10/14/2011 for the course MATH 100 taught by Professor A during the Fall '10 term at Wilfred Laurier University .

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Ch5_Radicals - MA100 Fall 2010 PART 1 Precalculus[5...

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