Radical expressions

Lecture Notes Radical Expressions page 1

Sample Problems

1. Simplify each of the following expressions.
p p p
a) 32 g) 7+2 7 2
p p 2
b) 45 h) 7 2
p p 3
c) 48x5 y 3 i) 3 1
p p p p p
d) 125 3 80 + 45 j) 5x 2 5x + 3
p
24 p p
e) p k) (2 x) (3 + 2 x)
54
p p p p p 2
f) 80a11 2 180a11 + 3 245a11 l) x 2

2. Rationalize the denominator in each of the following expressions.
p p p
3 1 2 5 1 8 5
a) p b) p c) p d) p e) p p
5 10 3 7+1 5 2 8+ 5
p p
3 5 3+ 5 p p p pp p
a) p + p d) 5 3 + 59 75 59
3+ 5 3 5
8 12 p p p p p p 2
b) p p +p p 5 7 3 e) 6+ 11 + 6 11
7+ 3 7 3
pp p pp p
c) 41 + 4 2 41 32
p
4. Find the exact value of x2 4x + 6 if x = 2 3.

5. Solve each of the following quadratic equations by completing the square. Check your solution(s).
a) x2 = 4x + 1 b) x2 + 13 = 8x

Practice Problems
p p p
a) 20 e) 5+2 5 2
p p p p 2
b) 300 2 75 + 12 f) 5 2
p p p p 3
c) 3 20 4 45 + 2 245 g) 5 2
p p p p p
d) 5 18a5 7 32a5 + 4 50a5 h) 3 5 1 7 5 + 2

c copyright Hidegkuti, Powell, 2008 Last revised: March 10, 2009


p
4 1 5 1 5x + 5
a) p c) p e) p g) p
7 10 + 3 5+1 x+1
p p
1 2 10 + 2 2
b) p d) p f) p p h) p
7 3 17 + 4 10 2 x+4
3. Find the exact value of
p p p
a) x2 6x + 1 if x = 3 10 b) b2 + 8b 20 if b = 5 2 c) x2 10x + 16 if x = 6+5

a) x2 + 47 = 14x b) x2 + 12x + 31 = 0 c) x2 = 2x + 1

Sample Problems – Answers

p p p p 2 p
1. a) 4 2 c) 4x2 y 3xy
b) 3 5 d) 4 5 e) f) 13a5 5a g) 3
3
p p p p p
h) 11 4 7 i) 10 + 6 3 j) 5x + 5x 6 k) 6 + x 2x l) x 2 2x + 2
p p p
3 5 p 1+ 7 p 13 4 10
2. a) b) 3 + 10 c) d) 3 + 5 e)
5 3 3
3. a) 7 b) 172 c) 3 d) 4 e) 22

4. 5
p p p p
5. a) 2 5; 2 + 5 b) 4 3; 4 + 3

Practice Problems – Answers

p p p p p p p
1. a) 2 5 b) 2 3 c) 8 5 d) 7a2 2a e) 1 f) 9 4 5 g) 17 5 38 h) 103 5
p p
4 7 3+ 7 p p p
2. a) b) c) 3 + 10 d) 8 + 2 17 or 2 4+ 17
7 2
p p p
3 5 5+3 p 2 x 8
e) f) g) 5 x + 5 h)
2 2 x 16
p
3. a) 2 b) 27 + 4 5 c) 3
p p p p p p
4. a) 7 2; 7 + 2 b) 6 5; 6+ 5 c) 1 2; 1 + 2



Sample Problems – Solutions
p p
(a) 32 = 4 2
Solution: We …rst factor the number under the square root sign into two factors, where the
…rst factor is the largest square we can …nd in the number. Then this …rst part can come out
from under the square root sign and become a coe¢ cient.
p p p p p
32 = 16 2 = 16 2 = 4 2
p p
(b) 45 = 3 5
Solution: We …rst factor the number under the square root sign into two factors, where the
…rst factor is the largest square we can …nd in the number. Then this …rst part can come out
from under the square root sign and become a coe¢ cient.
p p p p p
45 = 9 5 = 9 5 = 3 5
p p
(c) 48x5 y 3 = 4x2 y 3xy
Solution: We …rst factor the expressions under the square root sign into two factors, where
the …rst factor is the largest square we can …nd in the expression. Then this …rst part can
come out from under the square root sign.
p p p p p
48x5 y 3 = 16x4 y 2 3xy = 16x4 y 2 3xy = 4x2 y 3xy
p p p p
(d) 125 3 80 + 45 = 4 5
Solution:
p p p p p p
125 3 80 + 45 = 25 5 3 16 5 + 9 5
p p p p p p
= 25 5 3 16 5 + 9 5
p p p
= 5 5 3 4 5+3 5
p p p
= 5 5 12 5 + 3 5
p p
= (5 12 + 3) 5 = 4 5
p
24 2
(e) p =
54 3
Solution: We simlify both radical expressions and simplify.
p p p p p
24 4 6 4 6 2 6 2
p =p =p p = p =
54 9 6 9 6 3 6 3
p p p p
(f) 80a11 2 180a11 + 3 245a11 = 13a5 5a
Solution:
p p p
80a11 2 180a11 + 3 245a11 =
p p p
16a10 5a 2 36a10 5a + 3 49a10 5a =
p p p p p p
16a10 5a 2 36a10 5a + 3 49a10 5a =
p p p
4a5 5a 2 6a5 5a + 3 7a5 5a =
p p p
4a5 5a 12a5 5a + 21a5 5a =
p p
(4 12 + 21) a5 5a = 13a5 5a



p p
(g) 7+2 7 2 = 3
Solution: p p p p p p
7+2 7 2 = 7 7 2 7+2 7 4=7 4=3
p 2 p
(h) 7 2 = 11 4 7
Solution:
p 2 pp
7 2 = 7 2 7 2
p p p p
= 7 7 2 7 2 7+4
p p
= 7 4 7 + 4 = 11 4 7
p 3 p
(i) 3 1 = 10 + 6 3
p 2 p
Solution: We will …rst work out 3 1 and then multiply the result by 3 1 .
p 3 p p p
3 1 = 3 1 3 1 3 1
p p p p p
= 3 3 1 3 1 3+1 3 1
p p
= 3 2 3+1 3 1
p p
= 4 2 3 3 1
p p p p
= 4 3 4 2 3 3+2 3
p p
= 4 3 4 2 3+2 3
p p
= 4 3 4 6+2 3
p
= 10 + 6 3
p p p
(j) 5x 2 5x + 3 = 5x + 5x 6
Solution:
p p p pp p p p
5x 2 5x + 3 = 5x 5x + 3 5x 2 5x 2 3 = 5x + 3 5x 2 5x 6
p
= 5x + 5x 6
p p p
(k) (2 x) (3 + 2 x) = 6 + x 2x
Solution: p p p p p p p
2 x 3 + 2 x = 6 + 4 x 3 x 2 x x = 6 + x 2x
p p 2 p
(l) x 2 = x 2 2x + 2
Solution:
p p 2 p p p p
x 2 = x 2 x 2 we FOIL
p p p p p p p p
= x x x 2 2 x+ 2 2
p p p
= x 2x 2x + 2 = x 2 2x + 2



p
3 3 5
(a) p =
5 5
Solution: p rationalize the denominator, we will multiply both the numerator and denomi-
To
nator by 5. p p
3 3 5 3 5
p =p p =
5 5 5 5
1 p
(b) p = 10 + 3
10 3
Solution: To rationalize the denominator, we will multiply both the numerator and denomi-
p
nator by the conjugate of the denominator, which is 10 + 3.
p p
1 1 10 + 3 10 + 3 p
p = p p = = 10 + 3
10 3 10 3 10 + 3 1
The denominator is 1 since
p p p p p p
10 3 10 + 3 = 10 10 + 3 10 3 10 9 = 10 9=1
p
2 7 1
(c) p =
7+1 3
Solution: To rationalize the denominator, we will multiply both the numerator and denomi-
p
nator by the conjugate of the denominator, which is 7 1
p p p p
2 2 7 1 2 7 1 2 7 1 7 1
p =p p = = =
7+1 7+1 7 1 7 1 6 3
p
5 1 p
(d) p = 3+ 5
5 2
Solution:
p p p p p p p
5 1 5+2 5 1 5+2 5+2 5 5 2 3+ 5 p
p p = p p = p p = =3+ 5
5 2 5+2 5 2 5+2 5+2 5 2 5 4 1
p p p
8 5 13 4 10
(e) p p =
8+ 5 3
Solution: We multiply the fraction by 1 as a fraction of whose both numerator and denominator
are the conjugate of the denominator.
p p p p p p p p p p p p
8 5 8 5 8 5 8 5 8 5 8 5
p p =p p 1= p p p p = p p p p
8+ 5 8+ 5 8+ 5 8 5 8+ 5 8 5
We FOIL out both numerator and denominator
p p p p p p p p p p p
8 8 8 5 5 8+ 5 5 8 40 40 + 5 13 2 40
p p p p p p p p = =
8 8 8 5+ 5 8 5 5 8 5 3
Note: although this answer is acceptable, the expression can be further simpli…ed.
p p p p p p
13 2 40 13 2 4 10 13 2 4 10 13 2 2 10 13 4 10
= = = =
3 3 3 3 3



p p
3 5 3+ 5
(a) p + p =7
3+ 5 3 5
Solution:
p p p 2 p 2 p p
3 5 3+ 5 3 5 + 3+ 5 14 6 5 + 14 + 6 5 28
p + p = p p = = =7
3+ 5 3 5 3+ 5 3 5 9 5 4
8 12 p p
(b) p p +p p 5 7 3 = 172
7+ 3 7 3
Solution:
8 12 p p
p p +p p 5 7 3 =
7+ 3 7 3
p p p p
8 7 3 + 12 7 + 3 p p
= p p p p 5 7 3
7+ 3 7 3
p p p p
8 7 8 3 + 12 7 + 12 3 p p
= 5 7 3
p p 7 3
20 7 + 4 3 p p p p p p
= 5 7 3 = 5 7+ 3 5 7 3
4
= 25 (7) 3 = 175 3 = 172
pp p pp p
(c) 41 + 4 2 41 32 = 3
Solution:
q q q q r
41 + 4 2 41 32 = 41 + 32 41 32 = 41 + 32 41 32
s
p 2 p 2 p p
= 41 32 = 41 32 = 9 = 3
p p p pp p
(d) 5 3 + 59 75 59 = 4
Solution:
q q q q r
5 3 + 59 75 59 = 75 + 59 75 59 = 75 + 59 75 59
s
p 2 p 2 p p
= 75 59 = 75 59 = 16 = 4

p p p p 2
(e) 6 + 11 + 6 11 = 22
Solution:
q q 2 q 2 q 2 q q
p p p p p p
6 + 11 + 6 11 = 6+ 11 + 6 11 + 2 6 + 11 6 11
r
p p p p
= 6+ 11 + 6 11 + 2 6 + 11 6 11
p p
= 12 + 2 36 11 = 12 + 2 25 = 12 + 2 5 = 22



p
4. Find the exact value of x2 4x + 6 if x = 2 3. 5
Solution: We work out x2 …rst.
p p
2 p
x2 = 2 3
= 2 3 2 3
p p p p
= 4 2 3 2 3+ 3 3
p p
= 4 4 3+3=7 4 3
p
Now we substitute x = 2 3 into x2 4x + 6:
p p 2
x2 4x + 6 = 2 3
4 2 3 +6=
p p
= 7 4 3 8+4 3+6
= 7 8+6=5

p p
(a) x2 = 4x + 1 2 5; 2 + 5
Solution: We factor by completing the square.

x2 = 4x + 1 reduce one side to zero
x2 4x 1 = 0 (x 2)2 = x2 4x + 4
x2 {z + 4 4 1 = 0
| 4x }
(x 2)2 5 = 0
p 2
(x 2)2 5 = 0
p p
x 2+ 5 x 2 5 = 0
p p
x1 = 2 5 x2 = 2 + 5
p
We check: if x = 2 5, then
p 2 p p p p p p p p
LHS = 2 5 = 2 5 2 5 =4 2 5 2 5+ 5 5=4 4 5+5=9 4 5
p p p
RHS = 4 2 5 +1=8 4 5+1=9 4 5
p
and if x = 2 + 5, then
p 2 p p p p p pp p
LHS = 2+ 5 = 2+ 5 2+
5 =4+2 5+2 5+ 5 5=4+4 5+5=9+4 5
p p p
RHS = 4 2 + 5 +1=8+4 5+1=9+4 5



p p
(b) x2 + 13 = 8x 4 3; 4 + 3
Solution: We factor by completing the square.

x2 8x + 13 = 0 (x 4)2 = x2 8x + 16
2
x
| 8x + 16 16 + 13 = 0
{z }
(x 4)2 3 = 0
p 2
(x 4)2 3 = 0
p p
x 4+ 3 x 4 3 = 0
p p
x1 = 4 3 x2 = 4 + 3
p
We check: if x = 4 3, then
p 2 p p p p p
LHS = 4 3 + 13 = 4 3 4 3 + 13 = 16 4 3 4 3 + 3 + 13 = 32 8 3
p p
RHS = 8 4 3 = 32 8 3
p
and if x = 4 + 3, then
p 2 p p p p p
LHS = 4+ 3 + 13 = 4 + 3 4+ 3 + 13 = 16 + 4 3 + 4 3 + 3 + 13 = 32 + 8 3
p p
RHS = 8 4 + 3 = 32 + 8 3


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