Stitz-Zeager_College_Algebra_e-book

1 with a 5 and d 4 we get the nth denominator by 2

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Unformatted text preview: ion as the weapon of choice to solve this system. From A + C = 0, we get A = −C ; from 4B + 4D = 0, we get B = −D. Substituting these into the remaining two equations, we get √ √ −2C 2 − D − 2C 2 + D = 8 √ √ −4C − 2D 2 + 4C − 2D 2 = 0 or √ −4C √2 = 8 −4D 2 = 0 √ √ We get C = − 2 so that A = −C = 2 and D = 0 which means B = −D = 0. Our final answer is √ √ 8x2 x2 x2 √ √ = − x4 + 16 x2 − 2x 2 + 4 x2 + 2 x 2 + 4 8.6 Partial Fraction Decomposition 529 As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. In the end, if you take only one thing away from this section, it’s that 8 8 8 = 2− −9 x 9 For if the above were true, this section wouldn’t be here. x2 530 8.6.1 Systems of Equations and Matrices Exercises 1. Find o...
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