How would you solve the following inequality algebraically? $\displaystyle 0.5(x + 2)^2 (x - 4) > 2x + 3$ Is it. . . $\displaystyle 0.5(x + 2)^2 (x - 4) - 2x + 3 > 0$ And use interval charts?
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Originally Posted by Macleef How would you solve the following inequality algebraically? $\displaystyle 0.5(x + 2)^2 (x - 4) > 2x + 3$ Is it. . . $\displaystyle 0.5(x + 2)^2 (x - 4) - 2x + 3 > 0$ And use interval charts? Multiply by 2 throughout the equation $\displaystyle (x + 2)^2 (x - 4) > 4x + 6 $ $\displaystyle (x^2 + 4x + 4) (x - 4) > 4x + 6 $ $\displaystyle x^3 - 4x^2 + 4x^2 - 16x + 4x - 16 -4x - 6 > 0 $ $\displaystyle x^3 - 16x - 22 > 0 $ x = -2,9 ; -1,66 ; 4,5
Last edited by janvdl; Jan 13th 2008 at 08:57 PM. Reason: Thanks for the spot mr_fantastic
Originally Posted by janvdl Multiply by 2 throughout the equation Mr F explains: To make life easier. $\displaystyle (x + 2)^2 (x - 4) > 4x + 6 $ Mr F corrects: $\displaystyle (x^2 + 4x + 4) (x - 4) > 4x$ + 6 $\displaystyle x^3 - 4x^2 + 4x^2 - 16x + 4x - 16 - 4x$ - 6 > 0 $\displaystyle x^3 - 16x$ - 22 > 0 The roots of this equation still aren't exactly nice ones to work with ..... Alterations and paint work by Mr F.
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