The Mathematics 11
Competency Test
Adding and Subtracting Square
Roots
We’ve already covered the addition and subtraction of numerical square roots in some detail.
The procedure for adding and subtracting square roots which may contain algebraic expressions
is more or less the same:

simplify the square root in each term of the expression

combine terms whose square roots are identical
We will illustrate this strategy with a number of simple examples.
Example 1: Simplify
5 y 6 y 8 y 2 y
.
solution:
The square roots in all four of the terms in this expression are identical, all being just
y . So,
we just collect the “like terms:”
5 y  6 y  8 y  2 y  y 5  6  8  2
 y 5
5 y
as the simplest form of the result.
Example 2: Simplify
5 8 x 3  3 18 x 5 .
solution:
Neither of the two square roots occurring here are in simplest form. So, to start, we must simplify
each term.
Since
8x3 = 23x3 = 222x2x
we can write
8x3 
22 2 x 2 x  2 2
x2
2x  2x 2x
Also
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 1 of 6
18x5 = 232(x2)2x
and so
 
18 x 5  2 32  x 2
2
x 
x  32
2
2
2x  3 x 2 2x
Thus

5 8 x 3  3 18 x 5   5  2 x 2 x
  3  3 x
2
2x

 10x 2x  9x 2 2x

 10 x  9 x 2

2x
You could leave the final answer as this last expression, or you could still do a bit of factoring to
get, as the most simplified form
5 8x 3  3 18x 5  x 10  9x  2x
Example 3: Simplify 9 x 18 x
2
 3 50 x 4 .
solution:
This example is very similar to Example 2 above, so you should use it as a practice problem. Try
to solve it yourself before looking at our solution, given below.
First, simplify the individual square roots where possible. Since
18x2 = 232x2
we get
18 x 2 
2 32 x 2 
32
x2
2  3x 2
and since
50x4 = 252(x2)2
we get
50 x 4 
2 52
x 
2 2

52
x 
2 2
2  5x 2
2
Thus
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 2 of 6



9 x 18 x 2  3 50 x 4   9 x  3 x 2   3  5 x 2
2

 27 x 2 2  15 x 2 2

 27x 2  15x 2

2  12x 2 2
as the final answer.
2x 
Example 4: Simplify
8x 
18 x 
32x .
solution:
The last three of the square roots here can be simplified slightly:
8x 
22 2 x  2 2 x
18 x 
32 2 x  3 2 x
32 x 
42 2 x  4 2 x
and
Thus,
2x 
8x 
18x 

32x 
2x  2 2x  3 2x  4 2x
2 x 1  2  3  4 
 10 2x
as the final result.
Example 5: Simplify 5
pq 3  4 p pq  6q 9 pq .
solution:
We have
pq 3 
q 2 pq  q pq
9 pq 
32 pq  3 pq
and
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 3 of 6
Thus

5 pq 3  4 p pq  6q 9 pq   5  q pq
  4p

pq   6q  3 pq

 5q pq  4 p pq  18q pq

pq  5q  4 p  18q 
  23q  4 p 
pq
as the final answer.
The method should be fairly obvious by now. Use the remaining three examples as practice
problems. Try to do them yourself before looking at our solutions.
Example 6: Simplify 3 12y
3
 6y
48 y .
solution:
First, we check to simplify each of the two square roots individually:
12y 3 
22 3 y 2 y 
22
y2
3 y  2y 3 y
and
48y 
42 3 y 
3y  4 3y
42
Thus
3 12y 3  6 y

48 y   3  2y 3 y
   6y   4
3y

 6y 3y  24y 3y

3y
 6y  24y   18y
3y
as the final answer.
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 4 of 6
Example 7: Simplify 7t
18s 5t  3s 8s 3t 5 .
2
solution:
s 
2 2
18s 5t 
2 32
8s 3t 5 
22 2 s 2 s
st  3s 2
2st
and
t 
2 2
t  2st 2
2st
So

7t 2 18s 5t  3s 8s 3t 5   7t 2  3s 2
2st
 21s 2t 2

  3s   2st
2st  6s 2t 2

2
2st

2st

2st 21s 2t 2  6s 2t 2  15s 2t 2 2st
as the final answer.
Example 8: Simplify 3 p
8q 3  6q 32 p3  7 50q  5 18 p .
solution:
Each square root must first be simplified:
8q 3 
32 p3 
22 2 q 2 q  2q 2q
42 2 p 2 p  4 p 2 p
50q 
52 2q  5 2q
18 p 
32 2 p  3 2 p
and
Therefore
3 p 8q 3  6q 32 p3  7 50q  5 18 p

  3 p  2q 2q
  6q   4p


2p   7  5 2q
  5  3
2p

 6pq 2q  24pq 2p  35 2q  15 2p
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 5 of 6
  6 pq  35 
2q  15  24 pq 
2p
This is as simple as we can get the expression.
David W. Sabo (2003)
Adding and Subtracting Square Roots
Page 6 of 6