# Chapter Seven 7.6

```§ 7.6
A radical equation is an equation in which the variable occurs in a square root,
cube root, or any higher root. In this section, you will learn how to solve radical
equations.
When the variable occurs in a square root, it is necessary to square both sides of
the equation. When you square both sides of an equation, sometimes extra
answers creep in, called extraneous roots. For example, consider the following
very simple original equation.
x2
x2  4
x  2
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 7.6
Square both sides.
Solve the equation. But -2 does not
work in the original equation. Then -2 is
an extraneous root. It’s like a hitchhiker
that we picked up when we squared
both sides. It works only in the squared
form and is not a root of the original.
Just to note then….
When you solve a rational equation and must raise both sides to
an even power, remember to check your roots. Throw out any
extra (extraneous) roots from your solution set.
Another thing that you should know is … if your equation contains
two or more square root expressions, you will need to isolate one
square root expression, square both sides, and then you may have
to repeat the process. You may have to square both sides twice.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.6
nth Roots
1) If necessary, arrange terms so that one radical is
isolated on one side of the equation.
2) Raise both sides of the equation to the nth power to
eliminate the nth root.
3) Solve the resulting equation. If this equation still
contains radicals, repeat steps 1 and 2.
4) Check all proposed solutions in the original equation.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.6
Check Point 1
Solve:
3x  4  8.
SOLUTION
1) Isolate a radical on one side.
3x  4  8.
2) Raise both sides to the nth power. Because n, the index, is
2, we square both sides.


3x  4  8
2
2
3x  4  64
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.6
Simplify.
CONTINUED
3) Solve the resulting equation.
3x  4  64
3x  60
x  20
This is the equation from
step 2.
Subtract 5 from both sides.
Divide both sides by 2.
4) Check the proposed solution in the original equation.
Check 10:
?
64  8
3x  4  8
8  8 true
320  4 ? 8
The solution is 20.
60  4 ? 8
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.6
The solution set is {20}.
Number 4
Solve:
3x  2  5  0.
SOLUTION
1) Isolate a radical on one side. The radical can be isolated by
adding 5 to both sides. We obtain
3x  2  5.
2) Raise both sides to the nth power.
3x  2  5
2
3x  2  25
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.6
Simplify.
3x  2  25
CONTINUED
3) Solve the resulting equation.
3x  27
x9
4) Check the proposed solution in the original equation.
Check 9:
3x  2  5  0
39  2 - 5 ? 0
25  5  0
?
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.6
00
true
The solution is 9.
The solution set is {9}.
EXAMPLE (Also number 6)
Solve:
2x  5  11  6.
SOLUTION
1) Isolate a radical on one side. The radical, 2 x  5 , can be
isolated by subtracting 11 from both sides. We obtain
2x  5  5.
2) Raise both sides to the nth power. Because n, the index, is
2, we square both sides.


2 x  5   5
2
2
2 x  5  25
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.6
Simplify.
CONTINUED
3) Solve the resulting equation.
This is the equation from
step 2.
Subtract 5 from both sides.
Divide both sides by 2.
2 x  5  25
2 x  20
x  10
4) Check the proposed solution in the original equation.
Check 10:
2 x  5  11  6
5  11 ? 6
210  5 11? 6
?
25  11  6
16  6 false
Therefore there is no
solution to the equation.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.6
EXAMPLE
Solve: 3x  3x  7  5.
SOLUTION
1) Isolate a radical on one side. The radical, 3x  7 , can be
isolated by subtracting 3x from both sides. We obtain
 3x  7  5  3x.
2) Raise both sides to the nth power. Because n, the index, is
2, we square both sides.

3x  7

2
  5  3 x 
2
3x  7  25  30x  9 x 2
Simplify. Use the special
2
formula  A  B  A2  2 AB  B2 .
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.6
CONTINUED
3) Solve the resulting equation. Because of the x 2 -term, the
resulting equation is a quadratic equation. We need to write this
quadratic equation in standard form. We can obtain zero on the
left side by subtracting 3x and 7 from both sides.
3x  7  9 x 2  30x  25
0  9 x 2  27x  18
Equation from step 2.
0  9 x 2  3x  2
Subtract 3x and 7 from both
sides.
Factor out the GCF, 9.
0  x 2  3x  2
Divide both sides by 9.
0  x  2x  1
Factor the right side.


Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.6
CONTINUED
x20
x  2
x 1  0
x  1
Set each factor equal to 0.
Solve for x.
4) Check the proposed solutions in the original equation.
Check -2:
Check -1:
3x  3x  7  5
3x  3x  7  5
?
3 2  3 2  7 
5
31  31  7 ? 5
 6  1 ? 5
 3  4 ? 5
 6  1 ? 5
 7  5
? 5
3 2 
 5  5
false
The solution is -1. The solution set is {-1}.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 7.6
true
Number 10
Solve:
2 x  1  x  7.
Do on board
SOLUTION
x  12
4) Check the proposed solution in the original equation.
Check 12:
2x  1  x  7
212 1 ? 12  7
?
25 
5
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.6
55
true
In Summary…
Solving Radical Equations Containing nth Roots
1.
2.
3.
4.
Isolate one radical on one side of the equation.
Raise both sides to the nth power
Solve the resulting equation
Check proposed solutions in the original equation.
Sometimes proposed solutions will work in the final
simplified form of the original equation, but will not
work in the original equation itself. These imposter roots
that sometimes slip in when we square both sides of an
equation are called extraneous roots.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.6
DONE
EXAMPLE
x  4  x  4  4.
Solve:
SOLUTION
1) Isolate a radical on one side. The radical, x  4 , can be
isolated by subtracting x  4 from both sides. We obtain
x  4  4  x  4.
2) Raise both sides to the nth power. Because n, the index, is
2, we square both sides.

 
2
x4  4 x4

2
x  4  16  8 x  4  x  4
Simplify. Use the special
2
formula  A  B  A2  2 AB  B2 .
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.6
CONTINUED
x  4  20  x  8 x  4
Combine like terms.
1) Isolate a radical on one side. The radical, x  4, can be
isolated by subtracting 20 + x from both sides and then dividing
both sides by -8. We obtain
3  x  4.
2) Raise both sides to the nth power. Because n, the index, is
2, we square both sides.
3 
2

x4
9 x4

2
Square the 3 and the
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.6
x  4.
CONTINUED
3) Solve the resulting equation.
9 x4
This is the equation from the
last step.
5 x
Subtract 4 from both sides.
3) Check the proposed solution in the original equation.
Check 5:
x4  x4  4
5  4  5  4 ? 4
1  9 ? 4
44
true
The solution is 5. The
solution set is {5}.
1  3 ? 4
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.6
EXAMPLE
Solve: 2 x  3  7  10.
1
4
SOLUTION
Although we can rewrite the equation in radical form
4
2x  3  7  10,
it is not necessary to do so. Because, the equation involves a
fourth root, we isolate the radical term – that is, the term with
the rational exponent – and raise both sides to the 4th power.
2 x  3
1
4
 7  10
2 x  3
1
4
3
This is the given equation.
Subtract 7 from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.6
CONTINUED
4
2 x  3   34


2 x  3  81
1
4
2 x  78
x  39
Raise both sides to the 4th power.
Multiply exponents on the left
sides and then simplify.
Subtract 3 from both sides.
Divide both sides by 2.
Upon checking the proposed solution, 39, in the original
equation, we find that it checks out and is a solution. Therefore
the solution set is {39}.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.6
EXAMPLE
For each planet in our solar system, its year is the time it takes
the planet to revolve once around the sun. The function
f x   0.2 x
3
2
models the number of Earth days in a planet’s year, f (x), where x
is the average distance of the planet from the sun, in millions of
kilometers. Use the function to solve the following problem.
There are approximately 88 Earth days in the year of the planet
Mercury. What is the average distance of Mercury from the sun?
Use a calculator and round to the nearest million kilometers.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.6
CONTINUED
SOLUTION
To find the average distance of Mercury from the sun, replace
f (x) in the function with 88.
f x   0.2 x
88  0.2 x
440  x
3
2
3
2
 
440  x
2
3
2
3/ 2 2
193,600  x3
This is the given equation.
Replace f (x) with 88.
Divide both sides by 0.2.
Square both sides.
Simplify.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.6
CONTINUED
3
193,600  3 x 3
58  x
Take the cube root of both sides.
Simplify.
The model indicates that the average distance, to the nearest
million kilometers, that Mercury is from the sun is 58 million
kilometers.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 7.6
DONE
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