p143_26_2

p143_26_2 - Here is the details: First dividing 4 on both...

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Here is the details: First dividing 4 on both sides and moving 27(4 + β ) 2 on the right hand side: 4(6 + β ) 3 27(4 + β ) 2 4 ⇐⇒ (6 + β ) 3 27(4 + β ) 2 (1) Then we may do a sub u = 4 + β , then we get (6 + β ) 3 27(4 + β ) 2 ⇐⇒ ( u + 2) 3 27 u 2 (2) then expand ( u + 2) 3 = u 3 + 6 u 2 + 12 u + 8, then we get ( u + 2) 3 27 u 2 ⇐⇒ u 3 + 6 u 2 + 12 u + 8 27 u 2 (3) Then move 27 u 2 on the left hand side u 3 + 3 u 2 + 12 u + 8 27 u 2 ⇐⇒ u 3 - 21 u 2 + 12 u + 8 0 (4) Then our goal is to solve the cubic inequality(we may just take =, our goal is to find the smallest β ), there is no formula for us to solve it directly. So we usually guess a solution for this, it turns out that u = 1 is a solution, that means we find one factor (( u - 1)) for the left hand side of (5) u 3 - 21 u 2 + 12 u + 8 = 0 (5) Here we may write down u 3 - 21 u 2 + 12 u + 8 = ( u - 1)( au 2 + bu + c ) (6) here a,b,c are constants. You may expand right hand side of (7) and then compare the coef. of both
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This note was uploaded on 01/11/2012 for the course MATH 334 taught by Professor Dallon during the Fall '08 term at BYU.

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