# The all real solutions of equation x + 5 = 14 − 1 2 x . ### 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 8T

(a)

To determine

## To find: The all real solutions of equation x+5=14−12x .

Expert Solution

The real solution of equation x+5=1412x is 6.

### Explanation of Solution

Given:

The linear equation is x+5=1412x .

Calculation:

Simplify the given equation to find the value of x.

x+5=1412xx+514+12x=02x+1028+x2=0

Further simplify the above equation.

3x18=03x=18x=183x=6

Thus, the real solution of equation x+5=1412x is 6.

(b)

To determine

### To find: The all real solutions of expression 2xx+1=2x−1x .

Expert Solution

The real solution of expression 2xx+1=2x1x is 1.

### Explanation of Solution

Given:

The expression is 2xx+1=2x1x .

Calculation:

In the given expression eliminate the denominators to multiply each side by the lowest common denominator that is x(x+1) ,

(2xx+1)x(x+1)=(2x1x)x(x+1)2x2=(2x1)(x+1)2x2=2x2+2xx1x=1

Thus, the real solution of expression 2xx+1=2x1x is 1.

(c)

To determine

### To find: The all real solutions of equation x2−x−12=0 .

Expert Solution

The real solutions of equation x2x12=0 are 3 and 4.

### Explanation of Solution

Given:

The equation is x2x12=0 .

Calculation:

Factor the given equation to get the values of x,

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

Further solve above equation to get values,

(x+3)(x4)=0x=3,4

Thus, the real solutions of equation x2x12=0 are 3 and 4.

(d)

To determine

### To find: The all real solutions of equation 2x2+4x+1=0 .

Expert Solution

The real solutions of equation 2x2+4x+1=0 are 1±22 .

### Explanation of Solution

Given:

The equation is 2x2+4x+1=0 .

Calculation:

The Quadratic formula to find roots of equation ax2+bx+c=0 is,

x=b±b24ac2a

Where,

• a0

Substitute 2 for a, 4 for b and 1 for c from given equation to above formula,

x=4±(4)242122x=4±1684x=1±224x=1±22

Thus, the real solutions of equation 2x2+4x+1=0 are 1±22 .

(e)

To determine

### To find: The all real solutions of expression 3−x+5=2 .

Expert Solution

The expression 3x+5=2 has no real solution.

### Explanation of Solution

Given:

The expression is 3x+5=2 .

Calculation:

To eliminate the square root square both sides of given expression,

(3x+5)2=(2)23x+5=4

The above expression does not define any term of x to get the real solution so that there is no real solution for this expression.

Thus, the expression 3x+5=2 has no real solution.

(f)

To determine

### To find: The all real solutions of equation x4−3x2+2=0 .

Expert Solution

The real solutions of equation x43x2+2=0 are ±1 and ±2 .

### Explanation of Solution

Given:

The equation is x43x2+2=0 .

Calculation:

Factor the given equation to find the value of x.

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

Simplify the above expression.

x21=0,x22=0x2=1,x2=2x=±1,x=±2

Thus, the real solutions of equation x43x2+2=0 are ±1 and ±2 .

(g)

To determine

### To find: The all real solutions of equation 3|x−4|=10 .

Expert Solution

The real solutions of equation 3|x4|=10 are 23 and 223 .

### Explanation of Solution

Given:

The equation is 3|x4|=10 .

Calculation:

By the definition of absolute value 3|x4|=10 is equivalent to,

3(x4)=103x12=103x=22x=223

Or,

3(x4)=103x12=103x=2x=23

Thus, the real solutions of equation 3|x4|=10 are 23 and 223 .

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