Explanation Formula used: Quadratic formula : For a quadratic equation of the form a x 3 + b x + c = 0 the solutions are x = − b ± b 2 − 4 a c 2 a . Procedure used: Solving radical equations: 1. Isolate a single radical on one side of the equation. 2. Square both sides of the equation. (Or raise both sides to a power that is equal to the index of the radical.). 3. If a radica1 remains, repeat steps 1 and 2. 4. Solve the resulting equation. 5. All solutions must be checked in the original equation, and only those that satisfy the original equation are actual solutions. Calculation: The given equation can be written as, x − 2 − x − 1 = 30 1 x 2 − 1 x = 30 . Use the above procedure, to solve the equation 1 x 2 − 1 x = 30 . Simplify the above equation as follows. 1 x 2 − 1 x = 30 [ ∵ Isolating the radical ] 1 − x x 2 = 30 1 − x = 30 x 2 − 30 x 2 − x + 1 = 0 Substitute a = − 30 , b = − 1 and c = 1 in the quadratic formula to find the value of x . x = 1 ± ( − 1 ) 2 − 4 ( − 30 ) ( 1 ) 2 ( − 30 ) = 1 ± 1 + 120 − 60 = 1 ± 121 − 60 = 1 ± 11 − 60 On further simplification, x = − 1 + 11 60 or 1 − 11 60 = − 12 60 or 10 60 = − 1 5 or 1 6 Thus, the values of x are x = − 1 5 and x = 1 6

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ISBN: 9780135823002

Author: HARSHBARGER

Publisher: PEARSON

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Chapter 4.4, Problem 17E

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To solve: The equation x−2−x−1−30=0 algebraically and check the solution graphically or by substitution.

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Chapter 4.4, Problem 16E

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The equation x − 2 − x − 1 − 30 = 0 algebraically and check the solution graphically or by substitution.

Solution Summary: The author explains how to solve the equation x-2-X-1-30=0 algebraically and check the solution graphically or by substitution.

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To solve: The equation x−2−x−1−30=0 algebraically and check the solution graphically or by substitution.

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