Explanation Given information: The provided equation is 6 x 2 = 8 x + 5 . Formula used: If the equation is in the form of a x 2 + b x + c = 0 then its root is provide by the quadratic formula , x = − b ± b 2 − 4 a c 2 a Where, a is the coefficient of the x 2 term, b is the coefficient of the x term and c is the constant term. Calculation: Consider the equation, 6 x 2 = 8 x + 5 Rearrange the equation. 6 x 2 − 8 x − 5 = 0 The equation 6 x 2 − 8 x − 5 = 0 is in the form of a x 2 + b x + c = 0 . So, the values of a , b and c are, a = 6 b = − 8 c = − 5 Substitute a = 6 , b = − 8 and c = − 5 in the equation x = − b ± b 2 − 4 a c 2 a . x = − ( − 8 ) ± ( − 8 ) 2 − 4 ( 6 ) ( − 5 ) 2 ( 6 ) = 8 ± 64 + 120 12 = 8 ± 184 12 = 8 ± 13

Bundle: Elementary Technical Mathematics, Loose-leaf Version, 12th + WebAssign Printed Access Card, Single-Term

12th Edition

ISBN: 9781337890199

Author: Dale Ewen

Publisher: Cengage Learning

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Chapter 11, Problem 10T

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To calculate: The roots of the equation 6x2=8x+5 and round the result up to three significant digits, if required.

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Chapter 11, Problem 9T

Chapter 11, Problem 11T

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Chapter 11 Solutions

Bundle: Elementary Technical Mathematics, Loose-leaf Version, 12th + WebAssign Printed Access Card, Single-Term

Ch. 11.1 - Solve each equation: x2+x=12 Ch. 11.1 - Solve each equation: x23x+2=0 Ch. 11.1 - Solve each equation: x2+x20=0 Ch. 11.1 - Prob. 4E Ch. 11.1 - Solve each equation: x22=x Ch. 11.1 - Solve each equation: x215x=54 Ch. 11.1 - Solve each equation: x21=0 Ch. 11.1 - Solve each equation: 16n2=49 Ch. 11.1 - Solve each equation: x249=0 Ch. 11.1 - Prob. 10E

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The roots of the equation 6 x 2 = 8 x + 5 and round the result up to three significant digits, if required.

Solution Summary: The author calculates the roots of the equation 6x2=8x+5 and round the result up to three significant digits.

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To calculate: The roots of the equation 6x2=8x+5 and round the result up to three significant digits, if required.

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Chapter 11 Solutions