How Do We Solve Radical
Equations?
• Do Now: Simplify the given expression.
1.
2.
Radical Equations
An equation in which a variable occurs in the radicand
is called a radical equation. It should be noted, that
when solving a radical equation algebraically,
extraneous roots may be introduced when both sides of
an equation are squared. Therefore, you must check
your solutions for a radical equation.
Check: L.S.
R.S.
Solve: √ x - 3 - 3 = 0
x≥3
0
√x-3 -3
√x-3 =3
(√ x - 3 )2 = (3)2
√ 12 - 3 - 3
3-3
x-3=9
0
x = 12
Therefore, the solution
is x = 12.
If
2 x  1  2  5,
then x is equal to
(1) 1
(2) 2
(3) 5
(4) 4
2x  1  2  5
2x  1  3
2x  1  9
2 x  10
x5
What is the solution of the equation
2x  3  3  6 ?
(1) 42
(2) 39
(3) 3
(4) 6
2x  3  3  6
2x  3  9
2 x  3  81
2 x  84
x  42
Solving Radical Equations
4 + √ 4 + x2 = x
Check:
√ 4 + x2 = x - 4
(√ 4 + x2)2 = (x - 4)2
4 + x2 = x2 - 8x + 16
8x = 12
3
x
2
3 13
, the solution of
Since 
2
2
3
x=
is extraneous. Therefore,
2
there are no real roots.
4 4x
2
x
2
3

4  4  
2 
3
2
9
4 4
4
25
4
4
5
4
2
13
2
≠
3
2
The solution set of the equation
x6  x
is
(1) {–2,3}
(2) {–2}
(3) {3}
(4) { }
x6  x
x6 x
2
x  x6 0
2
( x  3)( x  2)  0
x  3 x  2
What is the solution set
of the equation
9 x  10  x ?
(1) {-1}
(2) {9}
(3) {10}
(4) {10, -1}
9 x  10  x
9 x  10  x
2
x  9 x  10  0
2
( x  10)( x  1)  0
x  10 x  1
x = -1 is an extraneous solution.
Solving Radical Equations
Solve
2x  4  x  7  0.
2x  4 

x7
2x  4    x  7 
2
2x  4  x  7
2(3)  4  3  7
10  10
0
Set up the equation so that
there will be one radical on
each side of the equal sign.
2
2x + 4 = x + 7
x=3
L.S.
x ≥ -2
Square both sides.
Simplify.
R.S.
0
Verify your solution.
Therefore, the
solution is
x = 3.
Squaring a Binomial
(a + 2)2 = a2 + 4a + 4
( 5 + √x - 2 )2
5 
x  2  2
10 x  2
5 
x 2
2
 25  10 x  2  (x  2)
 x  23 10 x  2
Note that the middle term is
twice the product of the two
terms of the binomial.
The middle term will
be twice the product
of the two terms.
A final concept that you
should know:
(a√x + b)2 = a2(x + b)
= a2x + ab
Solving Radical Equations
Solve
5x  1  3x  5  2.
5x 1  2  3x  5
 5x  1  2 
2
3x  5
Square both sides
of the equation.
5x 1  4  4 3x  5  (3x  5)
Use Foil.
5x 1  3x 1  4 3x  5
Simplify.
2x  2  4 3x  5
Simplify by dividing
by a common factor of 2.
x 1  2 3x  5
x  1  2 3x  5 
2
x  2x  1  4(3x  5)
2

2
Set up the equation
so that there will be
only one radical on
each side of the equal sign.
2
Square both sides of the
equation.
Use Foil.
Solving Radical Equations
x  2x  1  4(3x  5)
Distribute the 4.
x  2x 1 12x  20
Simplify.
2
2
x 10x  21  0
(x  3)(x  7)  0
x - 3 = 0 or x - 7 = 0
x = 3 or
x=7
2
L.S.
5x  1  3x  5
Factor the quadratic.
Solve for x.
Verify both solutions.
R.S.
L.S.
2
5x  1  3x  5
R.S.
5(3)  1  3(3)  5
5(7)  1  3(7)  5
42
64
2
2
2
One more to see another extraneous solution:
Thealgebraically
radical is already
but DOESisolated
NOT
3xa solution
 1  that
x you
3 find
make a true statement when you substitute it back
2
2 Square both sides
into the equation.
You must square the whole side
NOT each term.
3x  1  x  3
This must be FOILed
3 x  1  x  6 x  9 You MUST check
answers
2 you have a quadratic these
Since
equation
(has an x2
x

9
x

8

0
term) get everything on one side = 0 and see if


3
1

1

1

3
you can factor this
3
8


8

3
x  8x 1  0
Doesn't
work!
It
checks!
x  8, x  1
52 52 Extraneous
2
Let's try another one:
2 x  1
1
3
2 x  1
1  0
-1
First isolate the radical
-1
1 3
3
 1
3
2x  1  1
-1
-1
2 x  2
x  1
Remember that the 1/3
Now
is same
a 1/3thing
powersince
meansitthe
as a cube
root.
power
this
means the
same as a cube root so
cube both sides
Now solve for x
Let's check this answer
21  1  1  0
0  0 It checks!
3
Graphing a Radical Function
Graph y 
x  2.
The domain is x > -2.
The range is y > 0.
Solving a Radical Equation Graphically
Solve
The solution will be the
intersection of the graph
x  3  3  0.
y
x33
and the graph of
y = 0.
The solution
is x = 12.
Check:
L.S.
x33
12  3  3
9 3
33
R.S.
0
Solving a Radical Equation Graphically
Solve
5x  1  3x  5  2.
The solution is
x = 3 or x = 7.
Solving Radical Inequalities
Solve 7x  3  2  3.
Note the radical
7x - 3 is defined only
3
when x  .
Find the values for which
the graph of
y  7x  3  2
7
is above the graph of
y = 3.
The graphs intersect at
x = 4.
x>4
Therefore, the solution
is x > 4.
Solving Radical Inequalities
Solve
x 1  3.
x > -1
The graphs intersect
at the point where
x = 8.
x ≥ -1 and x < 8
The solution is
-1 < x and x < 8.
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