Digital Lesson
Solving Radical
Equations
Equations containing variables within radical signs are called
radical equations.
x 3  3
x9  x  7
3
x 1  2  3
A solution to a radical equation is a real number which, when
substituted for the variable, gives a true equation.
Examples: 1. Show that 12 is a solution of x  3  3.
(12)  3  3  9  3 True.
2. Show that 0 is a solution of 3 x  1  2  3.
3 (0)
 1  2  3  3 1  2  3  1  2  3 True.
3. Show that x  3 has no solutions.
The
symbol indicates the positive or
principal square root of a number.
Since x must be positive, x  3 has no solutions.
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If a and b are real numbers, n is a positive integer, and a = b,
then
an = bn
This principle can be applied to solve radical equations.
x  2, then ( x ) 2  2 2  x = 4.
Examples: 1. If
2. If
3
x  5, then (3 x ) 3  53  x = 125.
The converse of the statement is true for odd n and false for
even n.
If n is odd and a n = b n then a = b.
If n is even and a n = b n then it is possible that a  b.
Example: (3)2 = 9 = (3)2 but, 3  3.
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To solve a radical equation containing one square root:
1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve for the variable.
4. Check the solutions.
Example: Solve x  3  5  0.
x 3 5  0
x3  5
x  3  25
x  22
Original equation
Isolate the square root.
Square both sides.
Solve for x.
(22)  3  5  25  5  5  5  0 Check. True
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Example: Solve 4 x  x  3.
4 x x3
4 x  x3
16 x  ( x  3) 2
16 x  x 2  6 x  9
Original equation
Isolate the square root.
Square both sides.
Solve for x.
x 2  10 x  9  0
( x  1)( x  9)  0
x  1 or x  9
Solutions
4 (1)  (1)  4  1  3
Check. True
4 (9)  (9)  4(3)  9  12  9  3
Check. True
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Example: Solve 2  3 3 x  3  x.
2  3 3x  3  x
2  3 3x  x  3
8(3 x)  ( x  3) 3
Original equation
Isolate the cube root.
Cube both sides.
24 x  x 3  9 x 2  27 x  27
x 3  9 x 2  3 x  27  0
x 2 ( x  9)  3( x  9)  0
( x 2  3)( x  9)  0
(x2 + 3) does not give a root because
the square root of a negative number
is not a real number.
x9
2  3 3 (9)  3 2  3 27  3  2(3)  3  9
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Expand the cube.
Simplify.
Group terms.
Factor.
Solution
Check. True
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To solve a radical equation containing two square roots:
1. Isolate one radical on one side of the equation.
2. Square both sides of the equation.
3. Isolate the other radical.
4. Solve the equation.
5. Check the solutions.
Example: Solve x  8  x  4.
x 8  4 x
x  8  (4  x ) 2
x  8  16  8 x  x
8 x 8
64 x  64
x 1
(1)  8  (1)  4
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Isolate one radical.
Square both sides.
Simplify.
Isolate the other radical.
Square both sides.
Solution
Check. True
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Example: Solve x  2 x  2  6 .
x2 x2 6
2 x2  6 x
4( x  2)  36  12 x  x 2
x 2  16 x  28  0
( x  2)( x  14)  0
x  2 or x  14
Original equation
Isolate the square root.
Square both sides.
Simplify.
Factor.
Possible solutions
(2)  2 (2)  2  2  2 4  2  2( 2)  6 Check 2. True
2 is a solution of the original radical equation.
(14)  2 (14)  2  14  2 16  14  8  22  6 Check 14. False
14 is not a solution of the original radical equation.
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8
Example: A ten-foot board leans against an 8-foot wall so that
the top end of the board is at the top of the wall.
How far must the bottom of the board be from the wall?
Let x be the distance from the bottom of the board to the wall.
Use the Pythagorean Theorem.
x 2  82  10 2
x 2  100  64  36
x  6 or x  6
10 ft. board
Simplify.
8 ft.wall
Possible solutions
x
Check. Both are solutions of the radical equation, but since
the distance from the bottom of the board to the wall must
be nonnegative, – 6 is not a solution of the problem.
The bottom of the board must be 6 feet from the wall.
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Example: The time T (in seconds) taken for a pendulum of
length L (in feet) to make one full swing, back and
forth, is given by the formula
L
T  2
.
32
To the nearest hundredth, how long is a pendulum which takes 2
seconds to complete one full swing?
(3.24)
L
Radical
equation
2  2
2
 1.99 Check. True.
32
32
L
2
A pendulum of length
Isolate the radical.

32 2
approximately 3.24 feet will
2
L 1
make one full swing in 2
Square both sides.
 
32   
seconds.
32
L  2  3.24 Solve for L.

(to the nearest hundredth)
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