Transcript 1.6: Other Types of Equations

1.6
OTHER TYPES OF EQUATIONS
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What You Should Learn
• Solve polynomial equations of degree three or
greater.
• Solve equations involving radicals.
• Solve equations involving fractions or absolute
values.
• Use polynomial equations and equations
involving radicals to model and solve real-life
problems.
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Polynomial Equations
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Polynomial Equations
Example 1 shows how to use factoring to solve a
polynomial equation, which is an equation that can be
written in the general form
anxn + an – 1 xn – 1 + . . . + a2x2 + a1x + a0 = 0.
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Example 1 – Solving a Polynomial Equation by Factoring
Solve 3x4 = 48x2.
Solution:
First write the polynomial equation in general form with
zero on one side, factor the other side, and then set each
factor equal to zero and solve.
3x4 = 48x2
Write original equation.
3x4 – 48x2 = 0
Write in general form.
3x2(x2 – 16) = 0
Factor out common factor.
3x2(x + 4)(x – 4) = 0
3x2 = 0
Write in factored form.
x=0
Set 1st factor equal to 0.
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Example 1 – Solution
x+4=0
x = –4
Set 2nd factor equal to 0.
x–4=0
x=4
Set 3rd factor equal to 0.
cont’d
You can check these solutions by substituting in the
original equation, as follows.
Check
3(0)4 = 48(0)2
0 checks.
3(–4)4 = 48(–4)2
–4 checks.
3(4)4 = 48(4)2
4 checks.
So, you can conclude that the solutions are x = 0, x = –4,
and x = 4.
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Polynomial Equations
Occasionally, mathematical models involve equations that
are of quadratic type. In general, an equation is of
quadratic type if it can be written in the form
au2 + bu + c = 0
where a  0 and u is an algebraic expression.
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Equations Involving Radicals
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Example 4 – Solving Equations Involving Radicals
a.
Original equation
Isolate radical.
2x + 7 = x2 + 4x + 4
Square each side.
0 = x2 + 2x – 3
Write in general form.
0 = (x + 3)(x – 1)
Factor.
x+3=0
x = –3
Set 1st factor equal to 0.
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Example 4 – Solving Equations Involving Radicals
x–1=0
x=1
Set 2nd factor equal to 0.
By checking these values, you can determine that the only
solution is x = 1.
b.
Original equation
Isolate
Square each side.
Combine like terms.
Isolate
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Example 4 – Solving Equations Involving Radicals
x2 – 6x + 9 = 4(x – 3)
Square each side.
x2 – 10x + 21 = 0
Write in general form.
(x – 3)(x – 7) = 0
Factor.
x–3=0
x=3
Set 1st factor equal to 0.
x–7=0
x=7
Set 2nd factor equal to 0.
The solutions are x = 3 and x = 7. Check these in the
original equation.
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Equations with Fractions or
Absolute Values
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Example 6 – Solving an Equation Involving Fractions
Solve
Solution:
For this equation, the least common denominator of the
three terms is x(x – 2), so you begin by multiplying each
term of the equation by this expression.
Write original equation.
Multiply each term by
the LCD.
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Example 6 – Solution
2(x – 2) = 3x – x(x – 2)
2x – 4 = –x2 + 5x
x2 – 3x – 4 = 0
cont’d
Simplify.
Simplify.
Write in general form.
(x – 4)(x + 1) = 0
Factor.
x–4=0
x=4
Set 1st factor equal to 0.
x+1=0
x = –1
Set 2nd factor equal to 0.
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Example 6 – Solution
Check x = 4
cont’d
Check x = –1
So, the solutions are x = 4 and x = –1.
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Equations with Fractions or Absolute Values
To solve an equation involving an absolute value,
remember that the expression inside the absolute value
signs can be positive or negative.
This results in two separate equations, each of which must
be solved. For instance, the equation
|x – 2| = 3
results in the two equations x – 2 = 3 and –(x – 2) = 3,
which implies that the equation has two solutions: x = 5,
x = –1.
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Applications
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Example 8 – Reduced Rates
A ski club chartered a bus for a ski trip at a cost of $480. In
an attempt to lower the bus fare per skier, the club invited
nonmembers to go along. After five nonmembers joined the
trip, the fare per skier decreased by $4.80. How many club
members are going on the trip?
Solution:
Begin the solution by creating a verbal model and
assigning labels.
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Example 8 – Solution
cont’d
Labels: Cost of trip = 480
(dollars)
Number of ski club members = x
(people)
Number of skiers = x + 5
(people)
Original cost per member =
(dollars per person)
Cost per skier =
(dollars per person)
Equation:
Write
a fraction.
as
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Example 8 – Solution
Equation: (480 – 4.8x)(x + 5) = 480x
480x + 2400 – 4.8x2 – 24x = 480x
–4.8x2 – 24x + 2400 = 0
x2 + 5x – 500 = 0
(x + 25)(x – 20) = 0
cont’d
Multiply each side by x.
Multiply.
Subtract 480x from each side.
Divide each side by –4.8.
Factor.
x + 25 = 0
x = –25
x – 20 = 0
x = 20
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Example 8 – Solution
cont’d
Choosing the positive value of x, you can conclude that
20 ski club members are going on the trip. Check this in the
original statement of the problem, as follows.
Substitute 20 for x.
(24 – 4.80)25 ≟ 480
480 = 480
Simplify.
20 checks.
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