Algebra by Harley Flanders; Justin J Price

By Harley Flanders; Justin J Price

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The left side is not a perfect square, but something similar is: x2 + 6x + 9. So add - - 3. OUedrettc Equetlone 5 to both sides: x2 + 6x + 9 = 5, (x + 3)2 = 5. Therefore x + 3 must be one of the two square roots of 5, that is, x + 3 = + v's. Hence x = -3 + v's or x = -3 - v's . In general, each expression x2 + bx is part of a perfect square, b2 b 2 = (x2 + bx) + 4. x+2 ( ) Examples: x2 - lOx is part of (x - 5)2 = x2 - lOx + 25, is part of (x + � )2 = x2 + 3x + � · x2 + 3x Given x2 - lOx, we complete the square by adding 25, and given x2 + 3x, we complete the square by adding � .

2 . . . . 3 Compute and express in lowest terms: -I a) I + - x (b) x2x ++ •; I x+3. ( ( (x xy )2 ) (yx22 y ) SOLUTION factors: (a) Multiply numerators and denominators, then cancel common 2 - 1) = xy(x - l )(x + I) = x(x + 1) ( (x x-y 1 )2 ) (yx22 -+ yI ) = (x -xy(xJ)2(y2 + y) (x - 1)2y(y + I) (x - l )(y + I) " (b) Invert the divisor and multiply: + 3) . + +3 +l +1 +3 = +I + 1) There are no common factors; the product is already in lowest terms. __:____ �; x2 x (a) Answer • ( x2x ) ( x x ) (x x(x2l)(x x(x + 1) (x l)(y + I) _ = (b) (x + l)(x + 3) .

By -6) 11 3 = Answer (a) . I 27u6 · (x3 ) 11 3 = . = . = x g 11 3 -61 3 y x 2y - 2 = xy z -. 2 (24x4y Bz 12z) 11 4 (24) 114(x4 ) ll4( y s ) 1/4(z l2 ) 1/4z l/4 I . . xy2 (b) 27 u 6 . (3u 2 ) - 3 = . . (c) 2 . . . 2xy2z 3 0 . 2xy 2z 3 0. 2 Express as a single radical: (a) \19 Vf = (b) Vr3S5 -vr,:zs · -­ 8. SOLUTION (a) Convert to fractional exponents: \19 v'f = = 3 21 33 -112 3 213-1/2 = 3116 = Polynomlele 31 {13 . (b) Answer (a) {13 (b) rs2 \IS. EXERCISES Compute: 64312 5. 6. (1,000,000)516 Express as a power of 2: CV2 )213 9.

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