# Thread: When to use absolute value in square root of a variable?

1. ## When to use absolute value in square root of a variable?

In the first example, the textbook says we need the absolute value symbol because x might be negative
$\displaystyle \sqrt{25x^2} = 5\left|x\right|.$

In the second example, the textbook doesn't include the absolute value symbol
$\displaystyle \sqrt{12x^3}=2x\sqrt{3x}.$

What's the difference? When do we use absolute value and when not?

2. ## Re: When to use absolute value in square root of a variable?

In the first example, the textbook says we need the absolute value symbol because x might be negative
$\displaystyle \sqrt{25x^2} = 5\left|x\right|.$

In the second example, the textbook doesn't include the absolute value symbol
$\displaystyle \sqrt{12x^3}=2x\sqrt{3sx}.$

What's the difference? When do we use absolute value and when not?
If memory serves the convention with the absolute value is only true for numbers. Using variables we would have $\displaystyle \sqrt{25x^2} = \sqrt{25} \sqrt{x^2} = 5 \cdot \pm x = \pm 5x$.

Note that in the case of $\displaystyle \sqrt{12x^3}$ the odd power on the x forces the condition that $\displaystyle x \geq 0$ so there is no need for the +/-.

-Dan

3. ## Re: When to use absolute value in square root of a variable?

Originally Posted by topsquark
If memory serves the convention with the absolute value is only true for numbers. Using variables we would have $\displaystyle \sqrt{25x^2} = \sqrt{25} \sqrt{x^2} = 5 \cdot \pm x = \pm 5x$.

Note that in the case of $\displaystyle \sqrt{12x^3}$ the odd power on the x forces the condition that $\displaystyle x \geq 0$ so there is no need for the +/-.

-Dan
Sorry Dan, but $\displaystyle \sqrt{25}\ne\pm5$ because $\displaystyle \sqrt{25}=5$.

However, $\displaystyle \sqrt{25x^2}$ is defined for $\displaystyle x=-3$ therefore $\displaystyle \sqrt{25x^2}=5|x|$

As you correct observed for the odd case no absolute values are needed.

4. ## Re: When to use absolute value in square root of a variable?

Yes, in order that $\displaystyle \sqrt{3x^3}$ exist (we are talking about real numbers here) x itself must non-negative so |x|= x.

For example, in the first problem, $\displaystyle \sqrt{25x^2}$, if x= 3 or -3, we would have $\displaystyle \sqrt{25(3)^2}= \sqrt{25(9)}= \sqrt{225}= 15$ as well as $\displaystyle \sqrt{25(-3)^2}= \sqrt{25(9)}= \sqrt{225}= 15$. To cover both possibilities we must say $\displaystyle \sqrt{25x^2}= 5|x|$.

But in the second problem, $\displaystyle \sqrt{12x^3}$, if x= -3, that would be $\displaystyle \sqrt{12(-3)^3}= \sqrt{12(-27)}= \sqrt{-324}$ which is not a real number.

5. ## Re: When to use absolute value in square root of a variable?

Originally Posted by Plato
Sorry Dan, but $\displaystyle \sqrt{25}\ne\pm5$ because $\displaystyle \sqrt{25}=5$.

However, $\displaystyle \sqrt{25x^2}$ is defined for $\displaystyle x=-3$ therefore $\displaystyle \sqrt{25x^2}=5|x|$

As you correct observed for the odd case no absolute values are needed.
Thanks for the catch.

-Dan

6. ## Re: When to use absolute value in square root of a variable?

(25)^1/2 = +/- 5
(25x^2)^1/2=+/- 5*abs value x
if x =-3
(25)*(-3)^2 =225
225^1/2 = +/- 15

7. ## Re: When to use absolute value in square root of a variable?

Originally Posted by bjhopper
(25)^1/2 = +/- 5
(25x^2)^1/2=+/- 5*abs value x
if x =-3
(25)*(-3)^2 =225
225^1/2 = +/- 15
That is simply incorrect!

$\displaystyle \sqrt{25}\ne\pm 5$ We have $\displaystyle \sqrt{25}=(25)^{\frac{1}{2}}=5$