Explanation Given: ( x 2 − 3 x ) 2 + 2 ( x 2 − 3 x ) − 24 = 0 Calculation: Graph of the equation ( x 2 − 3 x ) 2 + 2 ( x 2 − 3 x ) − 24 = 0 is shown below From the graph the solution set of ( x 2 − 3 x ) 2 + 2 ( x 2 − 3 x ) − 24 = 0 is { − 1 , 4 } . Plug x = − 1 in ( x 2 − 3 x ) 2 + 2 ( x 2 − 3 x ) − 24 = 0 , we get ( ( − 1 ) 2 − 3 ( − 1 ) ) 2 + 2 ( ( − 1 ) 2 − 3 ( − 1 ) ) − 24 = 0 16 + 8 − 24 = 0 0 = 0 This is true

7th Edition

ISBN: 8220101539229

Author: Blitzer

Publisher: Pearson Education (US)

See similar textbooks

Videos

Question

Chapter 8.4, Problem 57E

To determine

To calculate: Use the graphing utility, find the real solution of the equation (x2−3x)2+2(x2−3x)−24=0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.4, Problem 56E

Chapter 8.4, Problem 58E

Students have asked these similar questions

For Exercises 115–120, factor the expressions over the set of complex numbers. For assistance, consider these examples. • In Section R.3 we saw that some expressions factor over the set of integers. For example: x - 4 = (x + 2)(x – 2). • Some expressions factor over the set of irrational numbers. For example: - 5 = (x + V5)(x – V5). To factor an expression such as x + 4, we need to factor over the set of complex numbers. For example, verify that x + 4 = (x + 2i)(x – 2i). 115. а. х - 9 116. а. х? - 100 117. а. х - 64 b. x + 9 b. + 100 b. x + 64 118. а. х — 25 119. а. х— 3 120. а. х — 11 b. x + 25 b. x + 3 b. x + 11

In Exercises 126–129, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 126. Once a GCF is factored from 6y – 19y + 10y“, the remaining trinomial factor is prime. 127. One factor of 8y² – 51y + 18 is 8y – 3. 128. We can immediately tell that 6x? – 11xy – 10y? is prime because 11 is a prime number and the polynomial contains two variables. 129. A factor of 12x2 – 19xy + 5y² is 4x – y.

In Exercises 101–103, perform the indicated operations. 1 1 1 101. x" – 1 x" + 1 x2" – 1 (1-X- -X ) (1 – (1 – 102. (1 - x + 1) x + 2 x + 3 103. (x – y)-1 + (x – y)-2

Chapter 8 Solutions

EBK INTERMEDIATE ALGEBRA FOR COLLEGE ST

Ch. 8.1 - Check Point 1 Solve: . Ch. 8.1 - Prob. 2CP Ch. 8.1 - Prob. 3CP Ch. 8.1 - Prob. 4CP Ch. 8.1 - Prob. 5CP Ch. 8.1 - Prob. 6CP Ch. 8.1 - Prob. 7CP Ch. 8.1 - Prob. 8CP Ch. 8.1 - Prob. 9CP Ch. 8.1 - Prob. 10CP

Knowledge Booster

$Algebra$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.

Use the graphing utility, find the real solution of the equation ( x 2 − 3 x ) 2 + 2 ( x 2 − 3 x ) − 24 = 0 .

Solution Summary: The author explains how to calculate the real solution of the equation (x2-3x

Videos

To calculate: Use the graphing utility, find the real solution of the equation (x2−3x)2+2(x2−3x)−24=0.

Want to see the full answer?

Chapter 8 Solutions