1.6 Other Types of Equations

1.6 Other Types of Equations
Chapter 1 Equations and Inequalities

Concepts and Objectives
⚫ Objectives for this section are:
⚫ Solve polynomial equations by factoring
⚫ Solve equations with radicals and check the solutions
⚫ Solve equations with rational exponents
⚫ Solve equations that are quadratic in form
⚫ Solve absolute value equations

Solving by Factoring
⚫ Some polynomial equations of degree 3 or higher can be
solved by moving all terms to one side (thus setting the
equation equal to 0), factoring the result, and setting
each factor to 0.
⚫ Usually, we will be looking for the GCF of each term.
⚫ Example: Solve by factoring: =
4 2
3 27
x x

Solving by Factoring (cont.)
⚫ Example: Solve by factoring:
Since 3 is a factor on both sides, we can divide both
sides by 3 in order to simplify each side. Even though x2
is also a factor, we generally do not divide by variables.
=
4 2
3 27
x x
( )
4 2
4 2
4 2
2 2
3 27
9
9 0
9 0
x x
x x
x x
x x
=
=
− =
− =
( )( )
2
3 3 0
0, 3,3
x x x
x
+ − =
= −

Solving by Grouping
⚫ If a polynomial consists of 4 terms, we can sometimes
solve it by grouping.
⚫ Grouping procedures require factoring the first two
terms, and then factoring the last two terms. If the
factors in the parentheses are identical, then the
expression can be factored by grouping.

Solving by Grouping
⚫ Example: Solve
Solutions are ‒1, 3, ‒3
3 2
9 9 0
x x x
+ − − =
( ) ( )
( )( )
( )( )( )
3 2
2
2
9 9 0
1 9 1 0
1 9 0
1 3 3 0
x x x
x x x
x x
x x x
+ − − =
+ − + =
+ − =
+ − + =
difference of squares!

Rational Exponent Equations
⚫ Recall that a rational exponent indicates a power in the
numerator and a root in the denominator.
⚫ There are multiple ways of writing an expression with a
rational exponent:
⚫ If we are given an equation in which a variable is raised
to a rational exponent, the simplest way to remove the
exponent on x is by raising both sides of the equation to
a power that is the reciprocal of the exponent (flip the
fraction).
( ) ( )
/ 1/
m
m
m n n m
n n
a a a a
= = =

⚫ Example: Solve
The reciprocal of is , so
⚫ To enter this on your calculator (or Desmos), use the ›
key (called a “caret”).
5/4
32
x =
5
4
4
5
( ) ( )
4/5 4/5
5/4
32
16
x
x
=
=

⚫ Example: Solve
We can now use the zero-factor property:
3/4 1/2
3x x
=
( )
3/4 1/2
3/4 2/4
2/4 1/4
3 0
3 0
3 1 0
x x
x x
x x
− =
− =
− =

⚫ Example: Solve
Solutions are 0 and
3/4 1/2
3x x
=
2/4
0
0
x
x
=
=
1/4
1/4
1/4
4
3 1 0
3 1
1
3
1 1
3 81
x
x
x
x
− =
=
=
 
= =
 
 
1
81

Power Property
⚫ Note: This does not mean that every solution of Pn = Qn
is a solution of P = Q.
⚫ We use the power property to transform an equation
that is difficult to solve into one that can be solved more
easily. Whenever we change an equation, however, it is
essential to check all possible solutions in the original
equation.
If P and Q are algebraic expressions, then every
solution of the equation P = Q is also a solution of
the equation Pn = Qn, for any positive integer n.

Solving Radical Equations
⚫ Step 1 Isolate a radical on one side of the equation.
⚫ Step 2 Raise each side of the equation to a power that is
the same as the index of the radical to eliminate the
radical.
⚫ If the equation still contains a radical, repeat steps 1
and 2 until no radicals remain.
⚫ Step 3 Solve the resulting equation.
⚫ Step 4 Check each proposed solution in the original
equation.

Solving Radical Equations (cont.)
⚫ Example: Solve − + =
4 12 0
x x

⚫ Example: Solve − + =
4 12 0
x x
= +
4 12
x x
= +
2
4 12
x x
− − =
2
4 12 0
x x
( )( )
− + =
6 2 0
x x
= −
6, 2
x

⚫ Example: Solve
Check:
Solution: {6}
− + =
4 12 0
x x
4 12
x x
= +
2
4 12
x x
= +
2
4 12 0
x x
− − =
( )( )
6 2 0
x x
− + =
6, 2
x = −
( )
− + =
6 4 6 12 0
− =
6 36 0
− =
6 6 0
=
0 0
( )
− − − + =
2 4 2 12 0
− − =
2 4 0
− − =
2 2 0
− 
4 0

⚫ Example: Solve + − + =
3 1 4 1
x x

⚫ Example: Solve + − + =
3 1 4 1
x x
( )
2
4 1
x + +
( )
2
2
3 1 4 1
3 1 4 2 4 1
2 4 2 4
2 4
4 4 4
5 0
5 0
0, 5
x x
x x x
x x
x x
x x x
x x
x x
x
+ = + +
+ = + + + +
− = +
− = +
− + = +
− =
− =
=

⚫ Example: Solve
Check:
Solution: {5}
+ − + =
3 1 4 1
x x
( )
2
2
3 1 4 1
3 1 4 2 4 1
2 4 2 4
2 4
4 4 4
5 0
5 0
0, 5
x x
x x x
x x
x x
x x x
x x
x x
x
+ = + +
+ = + + + +
− = +
− = +
− + = +
− =
− =
=
( )+ − + =
3 0 1 0 4 1
− =
1 4 1
− =
1 2 1
− 
1 1
( )+ − + =
3 5 1 5 4 1
− =
16 9 1
− =
4 3 1
=
1 1

You can also use the table on your graphing calculator to
check your answers.
⚫ Enter the left side of original equation into Y1 and the
right side of the equation into Y2 (use y¡ to create a
radical)
⚫ Go to the table (ys) and go to your first x-value. If
it is a solution, Y1 should equal Y2. If it doesn’t, it’s not a
solution.

Quadratic in Form
⚫ An equation is said to be quadratic in form if it can be
written as
where a  0 and u is some algebraic expression.
⚫ To solve this type of equation, substitute u for the
algebraic expression, solve the quadratic expression for
u, and then set it equal to the algebraic expression and
solve for x. Because we are transforming the equation,
you will still need to check any proposed solutions against
the original equation.
+ + =
2
0
au bu c

Quadratic in Form (cont.)
⚫ Example: Solve ( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
⅔ is two times ⅓
This is what makes it
quadratic in form.

⚫ Example: Solve
Let . This makes our equation:
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( )
= −
1 3
1
u x
+ − =
2
12 0
u u
( )( )
+ − =
4 3 0
u u
= −4, 3
u

⚫ Example: Solve
Let . This makes our equation:
So, and
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( )
= −
1 3
1
u x
+ − =
2
12 0
u u
( )( )
+ − =
4 3 0
u u
= −4, 3
u
( )
( ) ( )
1 3
3
1/3 3
1 4
1 4
1 64
63
x
x
x
x
− = −
 
− = −
 
− = −
= −
( )
( ) ( )
1 3
3
1/3 3
1 3
1 3
1 27
28
x
x
x
x
− =
 
− =
 
− =
=

⚫ Example: Solve (cont.)
Now, we have to check our proposed solutions:
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( ) ( )
− − + − − − =
2 3 1 3
63 1 63 1 12 0
( ) ( )
− + − − =
2 3 1 3
64 64 12 0
( ) ( )
− + − − =
2 1
4 4 12 0
− − =
16 4 12 0
=
0 0

⚫ Example: Solve (cont.)
Solution: {–63, 28}
(as before, you can also use your calculator to check)
( ) ( )
− + − − =
2 3 1 3
1 1 12 0
x x
( ) ( )
− + − − =
2 3 1 3
28 1 28 1 12 0
( ) ( )
+ − =
2 3 1 3
27 27 12 0
( ) ( )
+ − =
2 1
3 3 12 0
+ − =
9 3 12 0
=
0 0

Absolute Value Equations
⚫ You should recall that the absolute value of a number a,
written |a|, gives the distance from a to 0 on a number
line.
⚫ By this definition, the equation |x| = 2 can be solved by
finding all real numbers at a distance of 2 units from 0.
Both of the numbers 2 and ‒2 satisfy this equation, so
the solution set is {‒2, 2}.

Absolute Value Equations (cont.)
⚫ The solution set for the equation must include
both a and –a.
⚫ Example: Solve
=
x a
− =
9 4 7
x

⚫ The solution set for the equation must include
both a and –a.
⚫ Example: Solve
The solution set is
=
x a
− =
9 4 7
x
− =
9 4 7
x − = −
9 4 7
x
− = −
4 2
x − = −
4 16
x
=
1
2
x = 4
x
or
 
 
 
1
,4
2

⚫ Example: Solve 51 4 15 0
x
− − =

⚫ Example: Solve
Before we do anything else, we have to isolate the
absolute value expression:
51 4 15 0
x
− − =
51 4 15
1 4 3
x
x
− =
− =
1 4 3
4 2
1
2
x
x
x
− =
− =
= −
or
1 4 3
4 4
1
x
x
x
− = −
− = −
=
1
, 1
2
 
−
 
 

For Next Class
⚫ Section 1.6 in MyMathLab
⚫ Quiz 1.6 in Canvas
⚫ Optional: Read section 2.1 in your textbook
Remember that the deadline for the homework and quiz is
Sunday at 11:59 pm!

1.6 Other Types of Equations

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