Explanation Definition used: General quadratic equation : “Given that a , b , and c are constant ( a ≠ 0 ) , the equation a x 2 + b x + c = 0 is called the general quadratic equation in x ”. Calculation: Consider the equation ( C + 1 ) − 2 3 + 5 ( C + 1 ) − 1 3 − 6 = 0 . In order to find the value of C , first rewrite the equation in quadratic form. Substitute y = ( C + 1 ) − 1 3 in ( C + 1 ) − 2 3 + 5 ( C + 1 ) − 1 3 − 6 = 0 and simplify the equation as follows. y 2 + 5 y − 6 = 0 y 2 + 6 y − y − 6 = 0 y ( y + 6 ) − 1 ( y + 6 ) = 0 ( y + 6 ) ( y − 1 ) = 0 On further simplification, y + 6 = 0 or y − 1 = 0 y = − 6 or y = 1 ( C + 1 ) − 1 3 = − 6 or ( C + 1 ) − 1 3 = 1 [ ∵ y = ( C + 1 ) − 1 3 ] ( ( C + 1 ) − 1 3 ) − 3 = ( − 6 ) − 3 or ( ( C + 1 ) − 1 3 ) − 3 = 1 − 3 On further simplification, C + 1 = − 1 216 or C + 1 = 1 C = − 1 216 − 1 or C = 1 − 1 C = − 217 216 or C = 0 Thus, the solutions of the equation are C = − 217 216 and C = 0

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 14.3, Problem 18E

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To solve: The equation (C+1)−23+5(C+1)−13−6=0 algebraically.

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Chapter 14.3, Problem 17E

Chapter 14.3, Problem 19E

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The equation ( C + 1 ) − 2 3 + 5 ( C + 1 ) − 1 3 − 6 = 0 algebraically.

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To solve: The equation (C+1)−23+5(C+1)−13−6=0 algebraically.

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