# x^2 = 4 versus x = 4 ^ (1/2)

#### avr5iron

##### New member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2

#### MarkFL

Staff member
1.) $\displaystyle x^2=4$

Now, using the square root property, we find:

$\displaystyle x=\pm\sqrt{4}=\pm2$

2.) $\displaystyle x=4^{\frac{1}{2}}=\sqrt{4}=2$

You see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive.

In the second equation, we simply have x equal to a positive value, so there is just that one solution.

#### Poly

##### Member
Can someone explain why the solution for x in x^2 = 4 is x = 2, -2
while the solution for x in x = 4 ^ (1/2) is 2
I remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \implies x = 4^{\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \implies x = \pm 4^{\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency.

#### Deveno

##### Well-known member
MHB Math Scholar
naively, one might think:

$x^2 = 4$

therefore:

$(x^2)^{\frac{1}{2}} = 4^{\frac{1}{2}}$

that is:

$x^{(2)\left(\frac{1}{2}\right)} = x = 2$.

and we know that if $x = -2$ we have $x^2 = 4$, so what gives?

in general, the rule:

$(a^b)^c = a^{bc}$

only holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$.

while it is true that:

$(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like:

$(-4)^{\frac{1}{2}}$

and what this means is, when we write:

$x^{\frac{1}{2}} = y$

we are already tacitly assuming $x > 0$.

you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here:

y = x^(1/2) - Wolfram|Alpha

the "orange lines" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary).

on a deeper level, what is happening is this:

the "squaring function" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as "losing information about where we started from". as a result, we can only "partially recover" our beginnings, by taking a square root (we know the size, but we can only guess at the sign).

the symbol $\pm$ in the answer to $x^2 = 4$ (that is: $x = \pm 2$) is the way we indicate this uncertainty.

however, the function $y = x^{\frac{1}{2}}$ is only defined for $x \geq 0$ (we only get "the top half of the parabola" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2.

this indicates a peculiarity of functions: they can "shrink" or "collapse" their domains, but they only give ONE output for ONE input, so they cannot always "reverse themselves".

#### soroban

##### Well-known member
Hello, avr5iron!

$\text{Can someone explain why the solution is }\pm2\,\text{ for }\,x^2 \:=\: 4$
. . $\text{while the solution is }2\,\text{ for }\,x \:=\: 4^{\frac{1}{2}}$

The first is a quadratic equation; it has two roots.

. . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad (x-2)(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$

The second is a linear equation; one root.

. . $x \:=\:4^{\frac{1}{2}} \:=\:\sqrt{4} \quad\Rightarrow\quad x \:=\:2$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have explained this to my students like this . . .

If they give us a square root,
. . we assume it has the principal (positive) value.

So that: .$\sqrt{9} \:=\:3$

If we introduce a square root,
. . then we take the responsibility for both values.

So that: $x^2 \,=\,9 \quad\Rightarrow\quad x \,=\,\pm\sqrt{9} \quad\Rightarrow\quad x \,=\,\pm3$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply
numbers but when you have $x^2=4 \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt{4}$ then $\pm x=2$ so we are actually performing the square
root property on a variable now the result is not unique since a variable might have mutl-
When someone asks for $\sqrt{4}$ , are they not asking for a value to be determined x , such that
$4 = x \cdot x$