1.6 Other Types of Equations
See p 162, check out the graphs and the corresponding functions.
(Review) Solving a Polynomial
Equation by Factoring
1. Set = 0.
2. Factor.
3. Set each factor equal to zero. (Apply the zeroproduct principle.)
4. Solve the resulting equations.
5. Check the solutions in the original equation.
How do you recognize a “polynomial equation”?
Text Example
Don’t look at notes, no need to write.
• Solve by factoring: 3x4 = 27x2.
Step 1 Move all terms to one side and
obtain zero on the other side. Subtract
27x2 from both sides
3x4 - x2 = 27x2 - 27x2
3x4 - 27x2 = 0
Step 2 Factor.
3x4 - 27x2 = 0
3x2(x2 - 9)= 0
Solution cont.
• Solve by factoring: 3x4 = 27x2.
Steps 3 and 4 Set each factor equal to zero and
solve each resulting equation.
3x2 = 0
or
x2 - 9 = 0
x2 = 0
x2 = 9
x = 0
x = 9
x=0
x = 3
Steps 5 check your solution
If time: p 160 #6
x  1 = 9x  9x
3
2
(Rvw.)Solving RADICAL Equations
“IRS Audit”
•
ISOLATE the radical
•
RAISE each side to the same power
(match the index)
•
SOLVE the resulting equation
•
“AUDIT” (check), especially if raised
to an EVEN power. (ex: \/x = -2)
Ex: Solve:
x -3 7 = 9
Isolate
Raise
Solve
“Audit”
Why do we need to know this? Ex: p162#111
y = 5000 100 - x
Ex: Solve:
Isolate
Raise
Solve
“Audit”
3 x  x -5 = 8
Solving Radical Equations of the
Form xm/n= k
•
Assume that m and n are positive integers,
m/n is in lowest terms, and k is a real
number.
1. ISOLATE the expression with the rational
exponent.
2. RAISE both sides of the equation to the
n/m power.
3. SOLVE the resulting equation.
Solving Radical Equations of the
Form xm/n= k cont.
If m is even:
If m is odd:
xm/n = k
xm/n = k
(xm/n) n/m = ± kn/m
(xm/n)n/m = kn/m
x = ±kn/m
x = kn/m
It is incorrect to insert the ± when the
numerator of the original exponent is odd.
An odd index has only one root.
4. “AUDIT” all proposed solutions in the
original equation to find out if they are
actual solutions or extraneous solutions.
Text Example (No need to write, don’t look at
notes.)
Solve: x2/3 - 3/4 = -1/2.
Isolate x2/3 by adding 3/4 to both sides of the
equation: x2/3 = 1/4.
Raise both sides to the 3/2 power: (x2/3)3/2 =
±(1/4)3/2. (The square root of ¼ is ½, cubed is
1/8) (Note: WE take the square (even) root,
so we use +/-)
3
2
x = ±1/8. (Do p 160 # 34: ( x  5) = 8
)
Equations That Are Quadratic in Form
Some equations that are not quadratic can be
written as quadratic equations using an appropriate
substitution. Here are some examples.
Given Equation
Substitution
New Equation
x4 – 8x2 – 9 = 0
or
(x2)2 – 8x2 – 9 = 0
5x2/3 + 11x1/3 + 2 = 0
or
5(x1/3)2 + 11x1/3 + 2 = 0
t = _______
t2 – 8t – 9 = 0
t = ________
5t2 + 11t + 2 = 0
Hint: put the trinomial in descending order, use the variable part of
the 2nd term as t. Square it to see if it works for the variable part of
the first term. *** Don’t forget to SUBSTITUTE BACK! ***
Do p 160 #
46
2 x - 7 x - 30 = 0
(Rvw.)Rewriting an Absolute Value
Equation without Absolute Value Bars
• If c is a positive real number and X represents
any algebraic expression, then |X| = c is
equivalent to X = c or X = -c.
• (Using numbers: |2| = 2, and |-2| = 2, so if |x| =
2, that means that x = 2 or x = -2)
•* IMPORTANT* To apply this, we must
ISOLATE the absolute value part before setting
up cases and dropping the bars.
Example (Don’t look at notes, no need to write.)
Solve:
Answer:
3x - 1 = 4
3x-1 = 4
3x = 5
x = 5/3
and
3x-1 = -4
3x = -3
x = -1
Do on board: Solve | 1 – x | -7 = - 4 and | x + 2|+7 = 3
(Solve:
)
3x  7
-1 = 6
2