For Problems 1 − 12 , solve each inequality and graph its solution set on a number line. (Objective 1) ( x + 2 ) ( 4 x − 3 ) ≤ 0

Q: Expert Answer

To determine To plot: The solution sets of the quadratic inequality ( x + 2 ) ( 4 x − 3 ) ≤ 0 on a number line by solving it. Explanation Approach: The property of factoring is, a b = 0 if and only if a = 0 or b = 0 . If x ≤ a is the solution of a inequality then, the solution set is ( − ∞ , a ] . If x ≥ b is the solution of a inequality then, the solution set is [ b , ∞ ) . If x < a is the solution of a inequality then, the solution set is ( − ∞ , a ) . If x > b is the solution of a inequality then, the solution set is ( b , ∞ ) . If a ≤ x ≤ b is the solution of a inequality then, the solution set is [ a , b ] . If a < x < b is the solution of a inequality then, the solution set is ( a , b ) . Calculation: Consider the given quadratic inequality. ( x + 2 ) ( 4 x − 3 ) ≤ 0 Consider the quadratic equation, ( x + 2 ) ( 4 x − 3 ) = 0 … ( 1 ) From the above stated property the factors of equation ( 1 ) are given as follows. x + 2 = 0 x = − 2 Or, 4 x − 3 = 0 x = 3 4 The number line gets divided into three intervals, i.e., the numbers less than or equals to − 2 , the numbers between − 2 and 3 4 , and the numbers greater than or equals to 3 4 . This is shown in figure (1), Figure (1) Test whether each interval is positive or negative by choosing the random value between the interval. Consider the interval, x ≤ − 2 . Choose, x = − 4 . For, x = − 4 ,

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Answer 1

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Intermediate Algebra

10th Edition

Jerome E. Kaufmann + 1 other

Publisher: Cengage Learning

ISBN: 9781285195728

Solutions

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Answer 2

BuyFind

Intermediate Algebra

10th Edition

Jerome E. Kaufmann + 1 other

Publisher: Cengage Learning

ISBN: 9781285195728

Solutions

Chapter

Section

Answer 3

Solutions

Chapter

Section

Answer 4

Chapter

Section

Answer 5

To determine

To plot:

The solution sets of the quadratic inequality (x+2)(4x−3)≤0 on a number line by solving it.

Explanation of Solution

Approach:

The property of factoring is, ab=0 if and only if a=0 or b=0.

If x≤a is the solution of a inequality then, the solution set is (−∞,a].

If x≥b is the solution of a inequality then, the solution set is [b,∞).

If x<a is the solution of a inequality then, the solution set is (−∞,a).

If x>b is the solution of a inequality then, the solution set is (b,∞).

If a≤x≤b is the solution of a inequality then, the solution set is [a,b].

If a<x<b is the solution of a inequality then, the solution set is (a,b).

Calculation:

Consider the given quadratic inequality.

(x+2)(4x−3)≤0

Consider the quadratic equation,

(x+2)(4x−3)=0…(1)

From the above stated property the factors of equation (1) are given as follows.

x+2=0x=−2

Or,

4x−3=0x=34

The number line gets divided into three intervals, i.e., the numbers less than or equals to −2, the numbers between −2 and 34, and the numbers greater than or equals to 34. This is shown in figure (1),

Figure (1)

Test whether each interval is positive or negative by choosing the random value between the interval.

Consider the interval, x≤−2.

Choose, x=−4.

For, x=−4,

For Problems 1 − 12 , solve each inequality and graph its solution set on a number line. (Objective 1) ( x + 2 ) ( 4 x − 3 ) ≤ 0

Intermediate Algebra

Solutions

Intermediate Algebra

For Problems 1 − 12 , solve each inequality and graph its solution set on a number line. (Objective 1)

( x + 2 ) ( 4 x − 3 ) ≤ 0

Explanation of Solution

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