# To solve the below inequality in terms of intervals and illustrate the solution set on the real number line - x 2 ≥ 5

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter A, Problem 20E
To determine

## To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -   x2≥5

Expert Solution

The solution of the inequality are x5 and x5 and the solution set on a real number line -

### Explanation of Solution

Given: Inequality: x25

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers.

Calculation:

Given: Inequality equation is x25

Simplifying the above inequality, we have:

x250

x2(5)20

(x+5)(x5)0

To solve the above inequalities, we need to find the different intervals for which the inequality gives a value greater than 0 .

When x<5 :

(x+5) is negative and (x5) is negative.

Thus, (x+5)(x5)>0

So, x<5 is one of the solutions.

When x=5 :

(x+5) is Zero and (x5) is negative.

Thus, (x+5)(x5)=0

So, x=5 is one of the solutions.

When 5<x<5 :

(x+5) is positive and (x5) is negative.

Thus, (x+5)(x5)<0

So, 5<x<5 is not one of the solutions

When x=5 :

(x+5) is Positive and (x5) is Zero.

Thus, (x+5)(x5)=0

So, x=5 is one of the solutions.

When x>5 :

(x+5) is Positive and (x5) is positive.

Thus, (x+5)(x5)>0

So, x>5 is one of the solutions.

Combining all the solutions, we have:

the solutionsare x5 and x5

Drawing the above inequality on a real number line, we have:

Conclusion:

Hence, the solution of the inequality are x5 and x5 andthe solution set on a real number line -

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