Using completing square to simplify surds
Yue Kwok Choy
(1)
If
x  y   2
x > y > 0, then :
xy 
 x 2
2
x y
 y
2


x y

2
 x y
…. (1)
 3  2
2
Example : 5  24  3  2  2 6 
(2)
If we put
x + y = A , xy = B in
3 2
 2
2
 3  2
2

 3 2
(1) , then
A 2  x  y  x 2  2xy  y2
2



A 2  4B  x 2  2xy  y 2  4xy  x 2  2xy  y 2  x  y
A2  4B  x  y
Since x > y > 0 ,

x
2
x  y  x  y  A 
A 2  4B
2
2
Put all these in
and y 
x  y   x  y  A 
2
A 2  4B
2
(1) , that is,
x  y  2
xy  x  y
we have the result:
A > 0, B > 0 and A2 – 4B is a perfect square, then :
If
A  A 2  4B
A  A 2  4B

2
2
A2 B 
…. (2)
It is undesirable to memorize this formula, but rather understand the use of completing
square, as in the following example:
9  3  2
12  2 27
6  27 

2

(3)
If

9 3
2

2

9 3
2
93
2


2 9 3
2 2
 9  2
2

 3
A2 – 4B is not a perfect square, we can still use
9 3
2
 3
2
2 6
2
(2) , but we cannot simplify
expression.
82 6 
Example

8  8 2  46
8  8 2  46

2
2
8  40
8  40

2
2
, by (2)
 4  10  4  10
1
(4)
More examples :
(a)
y  74 3  74 3
Simplify
y
2
=
22  2 4  3  3  22  2 4  3  3
=
2  3 
2


2
2  3 
2
 

= 2 3  2 3  4
y = x  2 x 1  x  2 x 1
(b) Simplify
(i)
x  2,
If
y=

x 1  2
=

x 1 1 

=
x  1  1  12
2

2
 
x 1 1 




2
x 1  2
x  1  1  12

x 1 1
2

x 1 1  2 x 1
1x<2,
(ii) If
12  2
y=
1 
=

= 1
x  1  1  
x 1

2
 12  2
x  1  1  
x 1

2
  1  x  1
x  1   1  x  1  2
x 1
2
2
2 x  1
, if x  2
y
 2
, if 1  x  2

Note : If
(5)

x < 1 , y is undefined.
Exercise :
Simplify :
(a)
(b)
(c)
Ans.
18  8 3
2 3  12  6 3
1
1
3
2
2
1
1
3
2
2
1
12  140
(a)

1
8  60
3 5
(b)

2
10  80
3 2
(c)
0
2