2.9 SQUARE ROOT and
Is
•
9 = ±3 or is
9 =3?
9 = ±3 because we have learnt that x
x × x = x or we use our knowledge of exponents:
On the one hand we intuitively feel that
is a number multiplied by itself, i.e.
1
1
1
x 2 × x 2 = ( x 2 )2 = x .
•
2
On the other hand we have learnt that the graph of y = 9 − x is a positive semi-
•
circle. So the one Y-intercept is at y = 9 = 3 .
Sometimes we believe both ideas without realising that there is a contradiction!
The problem with x lies in language usage; non-transparent symbols, confusion
between language and symbols, messy notation habits, and ignorance of typical
mathematical processes of defining ...
In the beginning was the word
The word root refers to (is a synonym for) the solution of an equation; e.g.
Find the roots of the equation: x2 + 1 = 10.
According to the Fundamental Theorem of Algebra:
An equation of the first degree has one root (solution)
An equation of the second degree (quadratic equation) has two roots (solutions)
An equation of the third degree (cubic equation) has three roots (solutions)
2
A number taken to the power 2 is said to be squared, so x is called “x squared.”
3
A number taken to the power 3 is said to be cubed, so x is called “x cubed.”
That is why we say that the roots of a quadratic equation are square roots!
That is why we say that the roots of a cubic equation are cube roots!
The words “the square root of 9” means the solution of the equation x2 = 9. Thus 9
have two square roots, namely 3 en –3!!!
A quadratic equation has two square roots! And a cubic equation has 3 cube roots!
Confused? Remember, we are so far just talking about the words “square root”, we
have not yet said anything about the x symbol!
Messy Notation Habits – two wrongs do not make a right!
Part of the confusions is often caused by messy notations, for example:
We know that x2 – 9 = 0 has two square roots.. We know that we can find the solution
by factoring. However when we start using the
symbol, we might write it incorrectly:
x2 = 9
⇔ x=
9
= ±3
This strengthens the idea that 9 = ±3 ! Make sure that you know exactly where the
notation mistakes were made here! You must realise how this wrong notation creates
and reinforces the misconception that 9 = ±3 .
104
It would be wonderful if the symbol
root of 9” (with value ± 3 ).
9 had the same meaning as the words “the square
Unfortunately the intuitive meaning 9 = ±3 leads to contradictions and it is necessary
to re-visit this definition in order to prevent contradictions and ambiguity.
Example 1:
a 6 if a = -1.
Calculate
Method 1:
Method 2:
a 6 = ( −1) 6 = 1 = ±1
a 6 = a 3 = (−1) 3 = −1
How can two valid methods lead to two different answers??
Example 2:
We often say, without thinking twice, that x 2 = x .
With direct substitution of e.g. x = 3, this gives:
9 = 32 = 3 .
But we also find with direct substitution of e.g. x = –3 that:
Is it possible that both could be correct ?
9 = (−3) 2 = −3 ??
Example 3:
m n
Calculate (a ) if a = –3, m = 2 and n = ½.
Method 1:
(a ) = a
m n
mn
Method 2:
= (−3)
2× 12
1
= (−3) = −3
1
(a m ) n = [(−3) 2 ] 2 = 9 2 = 9 = ±3
1
How is it possible that two valid methods give two different answers??
Re-defining
These contradictions have to be somehow corrected. Mathematicians do this by returning to
the definition and changing the definition in such a way as to prevent later contradictions.
In the case of the square root symbol mathematicians re-defined it to be only positive. In
other words we define that 9 means only +3. It is therefore necessary to abort our
intuitive definition of “a number multiplied by himself” !!
We define x 2 = x , in other words
x 2 = x if x ≥ 0, or x 2 = − x if x < 0
Let’s return to the above examples and check that the contradictions have been eliminated:
Example 1:
Method 1:
Method 2:
a 6 = ( −1) 6 = 1 = | 1 | = 1
a 6 = | a 3 | = | (−1)3 | = | −1 | = 1
Example 2:
9 = 32 = | 3 | = 3
9 = (−3) 2 = | −3 | = 3
105
Example 3:
Method 1:
Method 2:
(a m ) n = a mn = (−3)
2× 12
1
= (−3)1 = −3
1
(a m ) n = [(−3) 2 ] 2 = 9 2 = 9 = | 3 | = 3
The first two contradictions have been solved, but not the third! We return to it below ...
Summary: Words and symbols
Let us make sure: 9 has two square roots, namely 3 and –3.
We may also write the result using the symbols 9 and − 9 .
9 = 3.
We define the symbol 9 as the principal1, or the positive square root of 9, so
We define the symbol − 9 as the negative square root of 9, so − 9 = −3 .
If x 2 = 9
2
y = 9− x
⇔ x = 9 = 3 or x = − 9 = −3
y = − 9 − x2
Condensed : x 2 = 9 ⇔ x = ± 9 = ±3
Note: Develop the habit to always read 9 as the “positive square root of 9”, not as the
“square root of 9”!! 9 is NOT the square root of 9!!
m n
mn
The exponential laws are also problematic – is (a ) = a ?
Example 4: Solve for x if x = 9 .
2
Method 1:
Method 2:
x =9
x2 = 9
2
1
1
⇔ ( x 2 ) 2 = (32 ) 2
⇔x=3
... because (a m ) n = a mn
⇔ x = ± 9 or ⇔ ( x + 3) ( x − 3) = 0
⇔ x = ±3
⇔ x = ±3
2
Example 5: Solve for x if x 3 = 9 .
Method 1:
Method 2:
2
3
2
x3 = 9
x =9
2
3
3
1
⇔ x = 33 .. .. because ( a m ) n = a mn
⇔ x = 27
1
( x 3 + 3)( x 3 − 3) = 0
1
⇔ (x 3 )2 = 9
⇔ ( x 3 ) 2 = (32 ) 2
1
⇔ x 3 = ± 9 = ±3
⇔ x = ±27
or
1
⇔ x 3 = ±3
x = ±27
You should check that in Examples 4 and 5 Method 2 gives the correct solutions. That
means that in Method 1 we have somewhere lost the logical equivalence!!!
To work properly with these exponential forms, we have to implement our “new”
definition of
a in Method 1 above. We have defined that
2
3
1
x 2 = | x | and this necessarily
1
1
3 2
2
also implies that ( x 2 ) 2 = | x | and even that ( x 3 ) 2 = ([( x 3 ) ] ) 2 = ( x ) 2 = | x | .
1
1
See http://mathworld.wolfram.com/SquareRoot.html
106
Example 4, Method 1:
Example 5, Method 1:
x =9
2
2
x3 = 9
1
1
⇔ ( x 2 ) 2 = (32 ) 2 ...... (1)
2
3
3
⇔ ( x 3 ) 2 = (32 ) 2 .......... ... (2)
⇔| x|=3
⇔ | x | = 33
⇔ x = ±27
⇔ x = ±3
1
The use of ( x ) 2 = | x | resolves our conflicts!
However, there is a danger that we will now write anything in sight as | x |! See above:
2
1
1
6
3 2
3
2
This is correct: ( x ) 2 = | x | and ( x 3 ) 2 = | x | and ( x ) 2 = [( x ) ] 2 = | x | (Example 1).
2
1
1
3
But ( x ) 3 ≠ | x | ! Surely, we want
3
3
1
1
1
− 8 = (−8) 3 = [(−2) 3 ] 3 = −2 !! So in this case: ( x 3 ) 3 = x !!
3
Example 6: Solve for x if x 2 = 8 .
3
x2 = 8
3
2
2
⇔ ( x 2 ) 3 = (23 ) 3
⇔ x = 2 2 .............. (3)
⇔ x =4
In (3) above we should not use | x | – check that x = –4 is not a root!
Confused? And where is the problem in these two contradictions?
1
2
1
1
2
1
− 2 ∉ R. But − 2 = ( −2) 2 = ( −2) 4 = 4 4 = [(2) 2 ] 4 = 2 4 = ( 2) 2 = 2 ∈ R ??
3
1
3
2
6
1
6
1
6
…….. (4)
1
6
− 8 = −2. But − 8 = (−8) = ( −8) = [( −8) ] = 64 = (2 ) = 2 ??
3
2
6
The problem obviously lies with the negative bases. That is why some mathematicians prefer to
p
q
limit the meaning of a and thus the validity of the exponential laws to cases where a ≥ 0 .
3
However, this would be extremely limiting. For example, x + 8 = 0 and x = 3 − 8 would
not be equivalent, and we need results like
p
q
3
− 8 = −2 !
1
1
1
a only has meaning if both a p and a q have meaning, to ensure that that (a p ) q = (a q ) p .
When a ≥ 0 there is no problem. But when a < 0 , it is only true for odd values of q, because
with real universe
q
1
a = a q , a < 0, q even is undefined e.g. − 4 , 4 − 16 , (−4) 2 ∉ R .
1
m n
mn
m n
mn
m
n
If a ≥ 0 , ( a ) = a . If a < 0 , ( a ) = a only if both a en a are defined!
3
2
3× 2
3
So, in (3) above: x is clearly positive (why?), so ( x 2 ) 3 = x 2
= x.
3
2
But in (2) above: x can be negative (why?2), so x does not exist , and so ( x 2 ) 3 =| x | .
3
2
This again emphasises: to maintain equivalence it is essential that we define the
Universal set (domain) precisely!
Can you now clarify the contradictions in (4)??
p
q
Click here for GSP or Excel graphs of y = x . Use the graphs to understand similar equations.
2
3
We need flexible thinking, e.g. think of (3) as
1
2
1
x 2 = ( x 2 )3 = ( x 3 ) 2
107
and think of (2) as
1
1
x 3 = ( x 3 )2 = ( x 2 ) 3
...