Chapter 0.4, Problem 10QR

a.

To determine

To find: the value of a for which the equation is true.

Answer to Problem 10QR

  $a\ge 0$

Explanation of Solution

Given information:

The equation:

  $\sqrt{a^2}=a$

Definition Used:

For a real number a ,

  $\sqrt{a^2}=\left|a\right|$

By the above definition,

  $\sqrt{a^2}=\left|a\right|$

Now, it is known that,

  $\left|a\right|=\left\{\begin{array}{l}a\text{ if }a\ge 0\\ -a\text{ if }a<0\end{array}\right.$

Case 1: $a\ge 0$

Then, $\left|a\right|=a$

Also the given equation is $\sqrt{a^2}=a$

So,

  $\begin{array}{c}\sqrt{a^2}=a\\ \left|a\right|=a\\ a=a\end{array}$

This case is always true.

Case 2: $a<0$

Then, $\left|a\right|=-a$

Also the given equation is $\sqrt{a^2}=a$

So,

  $\begin{array}{c}\sqrt{a^2}=a\\ \left|a\right|=a\\ -a=a\end{array}$

This case is never true.

Therefore, the given equation is true only for $a\ge 0$ .

b.

To determine

To find: the value of a for which the equation is true.

Answer to Problem 10QR

All real values of a

Explanation of Solution

Given information:

The equation:

  $\sqrt{a^2}=\pm a$

Definition Used:

For a real number a ,

  $\sqrt{a^2}=\left|a\right|$

By the above definition,

  $\sqrt{a^2}=\left|a\right|$

Also, the given equation $\sqrt{a^2}=\pm a$ is equivalent to:

  $\sqrt{a^2}=a$ or $\sqrt{a^2}=-a$

Now, it is known that,

  $\left|a\right|=\left\{\begin{array}{l}a\text{ if }a\ge 0\\ -a\text{ if }a<0\end{array}\right.$

Case 1: $a\ge 0$

Then, $\left|a\right|=a$

So,

  $\sqrt{a^2}=\left|a\right|=a$

Case 2: $a<0$

Then, $\left|a\right|=-a$

So,

  $\sqrt{a^2}=\left|a\right|=-a$

Thus, at least one of the two equations $\sqrt{a^2}=a$ or $\sqrt{a^2}=-a$ is true for all real values of a .

Therefore, the given equation is true for all real values of a .

c.

To determine

To find: the value of a for which the equation is true.

Answer to Problem 10QR

All real values of a .

Explanation of Solution

Given information:

The equation:

  $\sqrt{4a^2}=2\left|a\right|$

Definition Used:

For a real number a ,

  $\sqrt{a^2}=\left|a\right|$

Consider,

  $\begin{array}{c}\sqrt{4a^2}=\sqrt{4}\cdot \sqrt{a^2}\\ =2\sqrt{a^2}\\ =2\left|a\right|\end{array}$

So, the left hand side of given equation is equal to the right hand side of the equation for all values of a .

Therefore, the given equation is true for all real values of a .