Whether the statement “ ( 2 a − 3 ) 2 = 4 a 2 − 9 ” is true or false.

Question

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Answer 1

Question

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

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Answer 2

Question

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

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Answer 3

Question

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

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Answer 4

Question

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

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Answer 5

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

Answer 6

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

Answer 7

Chapter 6, Problem 1RE

To determine

Whether the statement “(2a−3)2=4a2−9” is true or false.

Answer 8

Expert Solution & Answer

Answer to Problem 1RE

The statement “(2a−3)2=4a2−9” is False_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Expand the expression (2a−3)2 as,

(2a−3)2=(2a)2−2(2a)(3)+(3)2 (∵(a−b)2=a2−2ab+b2)=4a2−12a+9

Thus, the expansion of (2a−3)2 is 4a2−12a+9.

Note that as the expansion of (2a−3)2 is 4a2−12a+9, the given statement cannot be true.

Therefore, the statement “(2a−3)2=4a2−9” is False_.

Answer 12

Ch. 6.1 - Prob. 1PE Ch. 6.1 - Prob. 2PE Ch. 6.1 - Prob. 3PE Ch. 6.1 - Prob. 4PE Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

Whether the statement “ ( 2 a − 3 ) 2 = 4 a 2 − 9 ” is true or false.

Whether the statement “(2a−3)2=4a2−9” is true or false.

Answer to Problem 1RE

Explanation of Solution

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Chapter 6 Solutions