To determine To calculate: The real solutions of the equation 3 x 4 + 8 x 2 − 1 = 0 . Explanation Given Information: The equation provided is 3 x 4 + 8 x 2 − 1 = 0 . Formula used: Steps to solve an equation in the quadratic form a u 2 + b u + c = 0 , a ≠ 0 where u is an algebraic expression. Step 1 : Substitute the variable of the equation with u to convert the equation in quadratic form a u 2 + b u + c = 0 where u is an algebraic expression. Step 2 : Solve the equation a u 2 + b u + c = 0 for u by factorizing. Step 3 : Replace u in the function with the original variable in step 1 . Step 4 : Now solve for the resulting equation for the original variable. Steps to determine the roots of the given quadratic equation, Step 1 : Compare the given quadratic equation with the standard form a x 2 + b x + c = 0 and note down the values of coefficients a , b and c . Step 2 : Substitute the value of coefficients in quadratic formula x = − b ± b 2 − 4 a c 2 a and solve for x . Zero factor property: It states that if a b = 0 , then a = 0 or b = 0 . Calculation: Consider the equation 3 x 4 + 8 x 2 − 1 = 0 , Let, u = n 2 Therefore, u 2 = n 4 Convert the equation 3 x 4 + 8 x 2 − 1 = 0 in quadratic form a u 2 + b u + c = 0 , by substituting these values in equation 3 x 4 + 8 x 2 − 1 = 0 , 3 u 2 + 8 u − 1 = 0 As the equation cannot be solved by factorizing therefore, the roots are determined by quadratic formula x = − b ± b 2 − 4 a c 2 a . On comparing the equation with the standard form a x 2 + b x + c = 0 , a ≠ 0 , a = 3 , b = 8 , c = − 1 Substitute these values in the quadratic formula, u = − b ± b 2 − 4 a c 2 a = − 8 ± ( 8 ) 2 − 4 ( 3 ) ( − 1 ) 2 ( 3 ) = − 8 ± 64 + 12 6 Further solve as, u = − 8 ± 76 6 = − 8 ± 2 19 6 = 2 ( − 4 ± 19 ) 6 = ( − 4 ± 19 ) 3 Hence u = − 4 ± 19 3 . Now substitute u with function of original variable x 2 , x 2 = − 4 ± 19 3 x = ± − 4 ± 19 3 Check the solution by substituting values of x obtained in the equation 3 x 4 + 8 x 2 − 1 = 0

Intermediate Algebra For College Students (10th Edition)

10th Edition

ISBN: 9780134758992

Author: Allen R. Angel, Dennis Runde

Publisher: PEARSON

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Chapter 8.4, Problem 80ES

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To calculate: The real solutions of the equation 3x4+8x2−1=0.

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Chapter 8.4, Problem 79ES

Chapter 8.4, Problem 81ES

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Here is an augmented matrix for a system of equations (three equations and three variables). Let the variables used be x, y, and z: 1 2 4 6 0 1 -1 3 0 0 1 4 Note: that this matrix is already in row echelon form. Your goal is to use this row echelon form to revert back to the equations that this represents, and then to ultimately solve the system of equations by finding x, y and z. Input your answer as a coordinate point: (x,y,z) with no spaces.

1 3 -4 In the following matrix perform the operation 2R1 + R2 → R2. -2 -1 6 After you have completed this, what numeric value is in the a22 position?

Chapter 8 Solutions

Intermediate Algebra For College Students (10th Edition)

Ch. 8.1 - Fill in the blanks with the appropriate word,...Ch. 8.1 - Fill in the blanks with the appropriate word,...Ch. 8.1 - Prob. 3ES Ch. 8.1 - Prob. 4ES Ch. 8.1 - Fill in the blanks with the appropriate word,...Ch. 8.1 - Prob. 6ES Ch. 8.1 - Prob. 7ES Ch. 8.1 - Prob. 8ES Ch. 8.1 - Prob. 9ES Ch. 8.1 - Prob. 10ES

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