Pre-Calculus 6.0
THE CONNECTION BETWEEN
SQUARE ROOTS AND ABSOLUTE VALUE
OBJECTIVE:
To clarify misconceptions about evaluating square roots, especially the
appropriate use of ± , as well as to firmly establish the connection between square
roots and absolute value.
KEY POINTS:
I. THE PRINCIPAL SQUARE ROOT OF A POSITIVE NUMBER IS
ALWAYS A POSITIVE NUMBER
From a graphical perspective:
Consider the graph of f ( x ) = x as shown to the right.
Notice that value of the function (i.e. the range values)
are always positive. The implication then is that the
square root of any positive value will always yield a
positive value. For example, 4 = 2 , not ± 2 as can be
seen from the graph. This root is called the principal
square root. It is only when you evaluate − 4 that the
answer is −2 ; this answer is known as the opposite of the
principal root.
y
4
f ( x) =
3
2
1
x
−1
1
2
3
4
−1
2
II. THE PRINCIPAL SQUARE ROOT OF A NEGATIVE NUMBER DOES NOT
EXIST IN THE REAL NUMBER SYSTEM.
The implication then is that the radicand (i.e. number or expression under the
radical sign) must always be non-negative (i.e. greater than or equal to zero). In
the case of a function where the radical is in the denominator, the radicand may
only be positive in value and cannot equal zero since division by zero is prohibited.
If the radicand is negative in value, the evaluation of that radical will require use
of the Complex number system which includes both Real number and Imaginary
number components (to be discussed later).
II. THE PRINCIPAL SQUARE ROOT OF A VARIABLE VALUE MUST YIELD A
VARIABLE EXPRESSION WHICH, IF EVALUATED, WOULD RESULT IN
ONLY POSITIVE VALUES.
Consider the expression x 2 . At first glance, it may seem reasonable to say that
x 2 . BUT, isn’t it also true
this expression simplifies to just “ x ” because ( x ) • ( x ) =
that ( − x ) • ( − x ) =x 2 too?
x
5
Pre-Calculus 6.0
THE CONNECTION BETWEEN
SQUARE ROOTS AND ABSOLUTE VALUE
II. (Cont’d)
How about a numerical example. Consider
( −5 )
2
. Again, at first glance, it may
seem reasonable to say that the answer is −5 . BUT, isn’t the answer actually
positive 5 because
( −5 )
2
=
25 = 5 which is supported by our graphical proof as
shown in Section I that the principal square root of any number is positive?
We can manage this confusion by considering the use of absolute value.
When the absolute value operation is applied to any number or expression, the
resulting number or expression is “forced” to be positive. As such,
=
52
( −5 )
2
=
5 5 and
=−5 =5
which now allows us to meet our requirement that the principal square root of any
number must be positive. In the same way for variable expressions,
x2 = x
( x + 2)
(2x
2
2
− 3 x − 5) =
2
=x + 2
2 x 2 − 3 x − 5 etc.
You can confirm these results on your graphing calculator. For example, enter
y1 =
x 2 into the equation editor on your calculator and graph. If the simplified
form of x 2 had been just “x”, you would expect to see a line with a slope of 1
going through the origin. BUT, as your calculator reveals, the actual graph is “V”
shaped and in fact represents the graph of y = x . Confirm by graphing
y 2 = x using the thick line style to see that these graphs are the same.