Explanation Given Information: The equation is x ( x − 5 ) = 2 . Formula used: The standard form of the quadratic equation is, a x 2 + b x + n = 0 Here, a , b , and n are the constants or number. Also, if the above quadratic equation is a perfect square. Then, n = [ 1 2 ( b ) ] 2 Here, b is the coefficient of x . Calculation: Consider the provided the equation, x ( x − 5 ) = 2 Simplify as, x 2 − 5 x = 2 On comparing x 2 − 5 x + N with the standard form of the quadratic equation a x 2 + b x + n , x 2 − 5 x + N is a complete square, for N = ( 1 2 b ) 2 . Here, b is the coefficient of x . Substitute − 5 for b , N = ( 1 2 ( − 5 ) ) 2 = ( − 5 2 ) 2 = 25 4 Thus, the value of N to complete the square for the equation x 2 − 5 x = 2 is 25 4 . Add 25 4 on the both sides of the equation, x 2 − 5 x = 2 . Hence, x 2 − 5 x + 25 4 = 2 + 25 4 x 2 − 5 x + 25 4 = 33 4 Factorize the equation for the calculating the complete square form of the equation. x 2 − 5 2 x − 5 2 x + 25 4 = 33 4 x ( x − 5 2 ) − 5 2 ( x − 5 2 ) = 33 4 ( x − 5 2 ) ( x − 5 2 ) = 33 4 ( x − 5 2 ) 2 = 33 4 Apply the square root property to simplify the equation. x − 5 2 = ± 33 4 x − 5 2 = ± 33 2 Add 5 2 on the both sides. x − 5 2 + 5 2 = ± 33 2 + 5 2 x = ± 33 2 + 5 2 Hence, x = 33 + 5 2 And, x = − 33 + 5 2 Hence, the solution for the equation is x = 33 + 5 2 , − 33 + 5 2

Beginning and Intermediate Algebra (6th Edition)

6th Edition

ISBN: 9780321969163

Author: Margaret L. Lial, John Hornsby, Terry McGinnis

Publisher: PEARSON

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Chapter 11.2, Problem 52E

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To calculate: The solution of the equation x(x−5)=2 by completing the square.

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Chapter 11.2, Problem 51E

Chapter 11.2, Problem 53E

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The solution of the equation x ( x − 5 ) = 2 by completing the square.

Solution Summary: The author calculates the solution of the equation x(x-5)=2 by completing the square.

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Chapter 11 Solutions