# The equation x + 5 = 14 − 1 2 x .

### 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

(a)

To determine

## To solve: The equation x+5=14−12x.

Expert Solution

The value of x is 6.

### Explanation of Solution

Consider the equation x+5=1412x.

Simplify the above expression as follows,

x+5=1412xx+12x=14532x=9x=9×23

On further simplification the above equation becomes,

x=3×2=6

Thus, the value of x is 6.

(b)

To determine

### To solve: The equation 2xx+1=2x−1x.

Expert Solution

The value of x is 1.

### Explanation of Solution

Consider the equation 2xx+1=2x1x.

Simplify the above expression as follows,

2xx+1=2x1x2x×x=(x+1)(2x1)2x2=2x2+2xx1x1=0

x=1

Thus, the value of x is 1.

(c)

To determine

### To solve: The equation x2−x−12=0.

Expert Solution

The values of x are 4,3.

### Explanation of Solution

Consider the equation x2x12=0.

Simplify the above expression as follows,

x2x12=0x24x+3x12=0x(x4)+3(x4)=0(x4)(x+3)=0

x=4,3

Thus, the values of x are 4,3.

(d)

To determine

### To solve: The equation 2x2+4x+1=0.

Expert Solution

The values of x are 1±122.

### Explanation of Solution

Formula used:

The value is x=b±b24ac2a.

Calculation:

Consider the equation x43x2+2=0.

Use the above mentioned Formula and simplifies the above expression as follows,

The value is

x=4±424×2×12×2=4±1684=4±84=4±224

=1±122

Thus, the values of x are 1±122.

Thus, the values of x are 1±122.

(e)

To determine

### To solve: The equation x4−3x2+2=0.

Expert Solution

The values of x are ±2,±1.

### Explanation of Solution

Consider the equation x43x2+2=0.

Simplify the above expression as follows,

x43x2+2=0x42x2x2+2=0x2(x22)1(x22)=0(x22)(x21)=0

Either,

x22=0x2=2x=±2

Or

x21=0x2=1x=±1

Thus, the values of x are ±2,±1.

(f)

To determine

### To solve: The equation 3|x−4|=10.

Expert Solution

The values of x are 223,23.

### Explanation of Solution

Consider the equation 3|x4|=10.

Simplify the above expression as follows,

3|x4|=10|x4|=103x4=±103

Take positive,

x4=103x=4+103=223

Take negative,

x4=103x=4103=23

Thus, the values of x are 223,23.

(g)

To determine

### To solve: The equation 2x(4−x)−12−34−x=0.

Expert Solution

The value of x is 125.

### Explanation of Solution

Consider the equation 2x(4x)1234x=0.

Simplify the above expression as follows,

2x(4x)1234x=0(4x)12(2x3(4x))=0(4x)12(2x12+3x)=0(4x)12(5x12)=0

On further simplification the above equation becomes,

(5x12)=05x=12x=125

Thus, the value of x is 125.

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