# The real solutions of the equation . | 2 x − 5 | = 9

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1, Problem 80RE
To determine

## To calculate: The real solutions of the equation . |2x−5|=9

Expert Solution

The solutions of the equation are x=7 and x=2 .

### Explanation of Solution

Given information:

The equation is given as |2x5|=9

Formula used:

In order to find all the solutions to higher-degree equation, use synthetic division, factoring, and the Quadratic Formula.

In order to Factorise the high degreepolynomial, determine all the terms that were multiplied together to get the given polynomial. Then try to factor each of the terms found in the first step. This continues until it can’t be factored anymore. When it can’t be factored further ,then polynomial is completely factored.

For an equation of the form ax2+bx+c=0 , the solution is given by quadratic formula given by:

x=b±b24ac2a .

Calculation:

Consider the equation |2x5|=9

This equation can be written as 2x5=9 and (2x5)=9 .

On solving 2x5=9 ,we get x=7 and on solving (2x5)=9 ,we get x=2 .

Thus, the real solutions of |x7|=4 are x=3 and x=11 .

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