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

• Nov 12th 2013, 10:59 AM
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
$\sqrt{25x^2} = 5\left|x\right|.$

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

What's the difference? When do we use absolute value and when not?
• Nov 12th 2013, 11:44 AM
topsquark
Re: When to use absolute value in square root of a variable?
Quote:

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

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

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 $\sqrt{25x^2} = \sqrt{25} \sqrt{x^2} = 5 \cdot \pm x = \pm 5x$.

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

-Dan
• Nov 12th 2013, 12:08 PM
Plato
Re: When to use absolute value in square root of a variable?
Quote:

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

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

-Dan

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

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

As you correct observed for the odd case no absolute values are needed.
• Nov 12th 2013, 01:06 PM
HallsofIvy
Re: When to use absolute value in square root of a variable?
Yes, in order that $\sqrt{3x^3}$ exist (we are talking about real numbers here) x itself must non-negative so |x|= x.

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

But in the second problem, $\sqrt{12x^3}$, if x= -3, that would be $\sqrt{12(-3)^3}= \sqrt{12(-27)}= \sqrt{-324}$ which is not a real number.
• Nov 12th 2013, 01:13 PM
topsquark
Re: When to use absolute value in square root of a variable?
Quote:

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

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

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

Thanks for the catch.

-Dan
• Nov 13th 2013, 03:35 AM
bjhopper
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
• Nov 13th 2013, 03:47 AM
Plato
Re: When to use absolute value in square root of a variable?
Quote:

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!

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