# To isolate: The radical in the equation 2 x + x = 0 .

### 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 1.5, Problem 4E

(a)

To determine

Expert Solution

## Answer to Problem 4E

The radical in the equation 2x+x=0 can be isolated as 2x=x_.

### Explanation of Solution

Consider the given equation 2x+x=0.

The term involving the radical sign is on the left-hand side of the equation.

Therefore, subtract x from both sides of the above equation to isolate the radical.

2x+xx=0x2x=x

Thus, the radical in the equation 2x+x=0 can be isolated as 2x=x_.

(b)

To determine

Expert Solution

## Answer to Problem 4E

Both sides of the equation 2x+x=0 can be squared as 2x=x2_.

### Explanation of Solution

Consider the equation 2x=x from part (a).

To solve this equation, the radical sign has to be eliminated.

Therefore, square both sides of the above equation.

(2x)2=(x)22x=x2

Thus, both sides of the equation 2x+x=0 can be squared as 2x=x2_.

(c)

To determine

Expert Solution

## Answer to Problem 4E

The solutions of the equation 2x=x2 are x=0or x=2_.

### Explanation of Solution

Formula used:

The solution of a quadratic equation of the form ax2+bx+c=0,a0 can be obtained by using the quadratic formula x=b±b24ac2a.

Calculation:

The equation 2x=x2 from part (b) can be rewritten as x22x=0.

Compare this equation with the general form ax2+bx+c=0.

Here a=1,b=2and c=0.

Substitute 1 for a, 2 for b and 0 for c in the quadratic formula to find the solution.

x=(2)±(2)24(1)(0)21=2±42=2±22=1+1or 11=2or 0

Thus, the solutions of the equation 2x=x2 are x=0or x=2_.

(d)

To determine

Expert Solution

## Answer to Problem 4E

The solution of the equation 2x=x2 that satisfy the original equation 2x+x=0 is x=0_.

### Explanation of Solution

From part (c) the solutions of the equation 2x=x2 are x=0or x=2.

To check whether these solutions satisfy the original equation, substitute these values in the equation 2x+x=0.

Substitute 0 for x in 2x+x=0.

2x+x=020+0=00=0[LHS=RHS]

Substitute 2 for x in 2x+x=0.

2x+x=022+2=04+2=04=0[LHSRHS]

The value x=2 does not satisfy the original equation. Therefore, it is not a solution of the equation 2x+x=0.

Thus, the solution of the equation 2x=x2 that satisfy the original equation 2x+x=0 is x=0_.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!