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: By using above procedure, the given equation can be solved as follows. x 2 5 − x 1 5 − 6 = 0 x 2 5 − x 1 5 = 6 [ ∵ Isolating the radical ] Let u = x 5 . Then the equation x 2 5 − x 1 5 − 6 = 0 becomes u 2 − u − 6 = 0 . Substitute a = 1 , b = − 1 and c = − 6 in the quadratic formula to find the value of u . u = 1 ± ( − 1 ) 2 − 4 ( 1 ) ( − 6 ) 2 ( 1 ) = 1 ± 1 + 24 2 = 1 ± 25 2 = 1 ± 5 2 On further simplification, u = 1 + 5 2 or 1 − 5 2 = 6 2 or − 4 2 = 3 or − 2 Substitute u = 3 in u = x 5 to find the value of x

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Author: HARSHBARGER

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Chapter 4.4, Problem 20E

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To solve: The equation x25−x15−6=0 algebraically; Check the solution graphically or by substitution.

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Chapter 4.4, Problem 19E

Chapter 4.4, Problem 21E

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Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The equation x 2 5 − x 1 5 − 6 = 0 algebraically; Check the solution graphically or by substitution.

Solution Summary: The author explains the procedure used to solve the equation x25- X. The solution is graphically or by substitution.

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To solve: The equation x25−x15−6=0 algebraically; Check the solution graphically or by substitution.

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