Chabot Mathematics
§7.6 2Var
Radical Eqns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Review § 7.6
MTH 55
 Any QUESTIONS About
• §7.6 → Radical Equations
 Any QUESTIONS About HomeWork
• §7.6 → HW-35
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Radical Equations
 A Radical Equation is an equation
in which at least one variable
appears in a radicand.
 Some Examples:
4
5 x 1  4  1 and m 2
4  1 and m 2  m  9.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Solve Eqns with 2+ Rad. Terms
1. Isolate one of the radical terms.
2. Use the Exponent Power Rule
3. If a radical remains, perform steps
(1) and (2) again.
4. Solve the resulting equation.
5. Check the possible solutions in
the original equation.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
x 3  x  2  1.
 SOLUTION
x 3  x  2  1
x 3  x  2  1

x 3
 
2

x 2 1
2
x  3   x 2   2 x 2  1
4  2 x 2
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
 SOLUTION 2  x 2
2 
2

x 2
x 3  x  2  1.

2
4 x2
6x
 Check 6 by Inspection → 3−2=1 
 Thus The number 6 checks
and it IS the solution
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve x 2  2  2 x3.

 SOLN
x 2
   2
2
x  2  4  4 2 x 3 
2 x 3


2
2 x 3

One radical is isolated.
We square both sides.
2
x  2  4  4 2 x 3  2 x  3
4
4 2 x 3  x  9
2 x 3

2
 ( x  9)
2
Square both sides.
16(2 x  3)  x  18 x  81
2
32 x  48  x  18 x  81
2
0  x  14 x  33
0  ( x  3)( x  11)
2
x  3 or
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x  11
Factoring
Using the principle of zero products
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve x 2  2  2 x3.
 Check: x = 3
x 2  2  2 x 3
x = 11
x 2  2  2 x 3
32  2  2(3)3
112  2  2(11)3
1  2  63
9  2  223
1  2  9
1  2  3
11

3  2  25
3  2  5
33

 The numbers 3 and 11 check and
are then confirmed as solutions.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
–x + 6 = 5.
x + 7 +
Start by isolating one radical on one side of the equation by subtracting
–x + 6 from each side. Then square both sides.
x + 7 +
–x + 6 = 5
x + 7 = 5 –
x + 7
x + 7
Twice the product of 5 and
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2
–x + 6
=
5 –
–x + 6
=
25 – 10
2
–x + 6 + (–x + 6)
–x + 6 .
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
x + 7 +
–x + 6 = 5.
This equation still contains a radical, so square both sides again.
Before doing this, isolate the radical term on the right.
x + 7
=
25 – 10
x + 7
=
31 – x – 10
2x – 24
=
–10
x – 12
=
–5
(x –
Chabot College Mathematics
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12)2
=
–5
–x + 6 + (–x + 6)
–x + 6
–x + 6
Subtract 31 and add x.
–x + 6
–x + 6
Divide by 2.
2
Square both sides again.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
–x + 6 = 5.
x + 7 +
This equation still contains a radical, so square both sides again.
(x –
x2
12)2
=
– 24x + 144 =
–5
(–5)2
–x + 6
2
–x + 6
Square both sides again.
2
(ab)2 = a2 b2
x2 – 24x + 144 =
25 ( –x + 6 )
x2 – 24x + 144 =
–25x + 150
Distributive property
x2 + x – 6 =
0
Standard form
(x + 3)(x – 2) =
0
Factor.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
x + 7 +
–x + 6 = 5.
Now finish solving the equation.
(x + 3)(x – 2) =
0
x + 3 = 0
or
x – 2 = 0
x = –3
or
x = 2
Zero-factor property
Finally CHECK for Extraneous Solutions
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Solve
x + 7 +
–x + 6 = 5.
Check each potential solution, –3 and 2, in the original equation.
If x = –3, then
x + 7 +
–3 + 7 +
If x = 2, then
–x + 6
= 5?
x + 7 +
–x + 6
–(–3) + 6 = 5?
2+ 7 +
–(2) + 6 = 5?
4 +

9 = 5?
5 = 5
9 +

The solution set is { −3, 2 }.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
= 5?
4 = 5?
5 = 5
The Principle of Square Roots
 Recall the definition of the
PRINCIPAL Square Root
 For any NONnegative real number
n, If x2 = n, then
x  n or x   n.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Recall The Pythagorean Theorem
 In any right triangle, if a and b are
the lengths of the legs and c is the
length of the hypotenuse, then
a2 + b2 = c2
Hypotenuse
c
b
Leg
a
Leg
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Pythagorus
 How long is a guy wire if it reaches from
the top of a 14 ft pole to a point on the
ground 8 ft from the pole?
 SOLUTION d  14  8  196  64  260
2
d
14
8
Diagram
2
2
• We now use the principle of
square roots. Since d represents
a length, it follows that d is the
positive square root of 260:
d  260 ft
d  16.125 ft.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Isosceles Right Triangle
 When both legs of a right triangle
are the same size, we call the
triangle an isosceles right triangle.
If one leg of an isosceles right
triangle has length a then
2
2
2
c a a
c  2a
2
c  2a
2
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c  a 2.
2
2
c
a
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
a
Lengths for Isosceles Rt Triangles
 The length of the hypotenuse in an
isosceles right triangle is the length
of a leg times 2.
45o
a 2
a
45o
a
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  Isosceles Rt. Tri.
 The hypotenuse of an isosceles right
triangle is 8 ft long. Find the length of a
leg. Give an exact answer and an
approximation to three decimal places.
 SOLUTION
8
a
a 2
8
=
2
45o
8
a
45o
a
a 2  a 2  82
2
2
2
a

8
Chabot College Mathematics
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4 2  a.
( after Rationalizing Divisor).
Exact answer: a  4 2 ft
Approximation: a  5.657 ft.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
30°-60°-90° Triangle
 A second special
triangle is known
as a 30°-60°-90°
right triangle, so
named because of
the measures of its
angles
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Lengths for 30/60/90 Rt Triangles
 The length of the longer leg in a
30/60/90 right triangle is the length
of the shorter leg times 3. The
hypotenuse is twice as long as the
shorter leg.
30o
2a
a 3
60o
a
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  30°-60°-90° Triangle
 The shorter leg of a 30/60/90 right
triangle measures 12 in. Find the
lengths of the other sides. Give exact
answers and, where appropriate, an
approximation to three decimal places.
 SOLUTION
• The hypotenuse is twice as long
as the shorter leg, so we have a 3
c = 2a = 2(12) = 24 in.
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30o
2a
60o
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Example  30°-60°-90° Triangle
 SOLUTION
 The length of the longer leg
is the length of the shorter
leg times 3. This yields
30o
2a
a 3
60o
12
b  a 3 = 12 3 in.
Exact answer: c  24 in., b  12 3 in.
Approximation: b  20.785 in.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
WhiteBoard Work
 Problems From §7.6 Exercise Set
• 24, 34, 38, 48, 60

Astronomical Unit
= Sun↔Earth
Distance
= 149 598 000 km
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
All Done for Today
The Solar
Star System
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-46_sec_7-6b_2Var_Radical_Eqns.ppt
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