Explanation Result used: Complex Conjugate Root Theorem: “If the coefficients of the equation f ( x ) = 0 and real and a + b j ( b ≠ 0 ) is a complex root, then its conjugate, a − b j , is also a root”. Calculation: The given equation ( 4 y 2 + 9 ) ( 25 y 2 − 10 y + 1 ) = 0 can be written as follows. ( 4 y 2 + 9 ) ( 25 y 2 − 5 y − 5 y + 1 ) = 0 ( 4 y 2 + 9 ) ( 5 y ( 5 y − 1 ) − 1 ( 5 y − 1 ) ) = 0 ( 4 y 2 + 9 ) ( 5 y − 1 ) 2 = 0 By inspection, substitute y = 0 in the equation ( 4 y 2 + 9 ) ( 5 y − 1 ) 2 = 0 . ( 4 ( 0 ) 2 + 9 ) ( 5 ( 0 ) − 1 ) 2 = ? 0 ( 9 ) ( − 1 ) 2 = ? 0 9 = 0 ( False ) Therefore, y = 0 is not a root of the equation ( 4 y 2 + 9 ) ( 5 y − 1 ) 2 = 0 . Substitute y = 1 5 in the equation ( 4 y 2 + 9 ) ( 5 y − 1 ) 2 = 0 . ( 4 ( 1 5 ) 2 + 9 ) ( 5 ( 1 5 ) − 1 ) 2 = ? 0 ( 4 25 + 9 ) ( 1 − 1 ) 2 = ? 0 0 = 0 ( True ) Therefore, y = 1 5 is double root of the equation ( 4 y 2 + 9 ) ( 5 y − 1 ) 2 = 0

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

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ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 15.2, Problem 6E

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The roots of the equation (4y2+9)(25y2−10y+1)=0 using inspection.

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Chapter 15.2, Problem 5E

Chapter 15.2, Problem 7E

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Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

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