Category Archives: Trigsted Chapter 1

1.6 Other Types of Equations, 5.1 Polynomial Functions and Models

Polynomial Equation (Solve by factoring with the grouping method)

Example:  Solve the polynomial equation

y^3+y^2=4y+4

Solution:  Solve the polynomial equation by factoring.

The original equation.
y^3+y^2=4y+4
Write the equation so that all the terms are on the same side.

  • Subtract 4y
  • Subtract 4
y^3+y^2=4y+4
y^3+y^2-4y=4y-4y+4
y^3+y^2-4y=4
y^3+y^2-4y-4=4-4
y^3+y^2-4y-4=0
Group two terms pairs of terms and factor the greatest common factor from each group.

  • The greatest common factor for the first group is y^2
  • The greatest common factor for the second group is -4
y^3+y^2-4y-4=0
(y^3+y^2)+(-4y-4)=0
(y^3+y^2)+(-4y-4)=0
y^2(y+1)+-4(y+1)=0
Factor the common binomial from each term.

  • The common binomial is (y+1)
  • The other factor is formed using the coefficients of the parenthesis
y^2(y+1)+-4(y+1)=0
(y+1)(y^2-4)=0
Continue to factor completely by factoring the difference of squares in the second parenthesis
(y+1)(y^2-4)=0
(y+1)(y+2)(y-2)=0
Apply the zero product property by setting each factor equal to zero.
y+1=0 or y+2=0 or y-2=0
Solve each remaining equation.
y+1=0 or y+2=0 or y-2=0
y+1-1=0-1 or y+2-2=0-2 or y-2+2=0+2
y=-1 or y=-2 or y=2

The solutions to the polynomial equation y^3+y^2=4y+4 are y=-1 or y=-2 or y=2.

1.4 Quadratic Equations

Solving a Quadratic Equation using the Quadratic Formula: Example 1 of 1

Example:  Solve the quadratic equation with the quadratic formula.

2x^2+7=4x

Solution:

The original equation.
2x^2+7=4x
Write the equation so that all of the terms are on the same side.
2x^2+7=4x
2x^2+7-4x=4x-4x
2x^2-4x+7=0
Identify a, b and c.
a=2, b=-4, c=7
2x^2-4x+7=0
The quadratic formula.
x={-b pm sqrt{b^2-4ac}}/{2a}
Substitute the values into the quadratic formula.
x={-b pm sqrt{b^2-4ac}}/{2a}
x={-(-4) pm sqrt{(-4)^2-4(2)(7)}}/{2(2)}
Simplify using order of operations by applying powers, multiplying and then subtracting.
x={-(-4) pm sqrt{(-4)^2-4(2)(7)}}/{2(2)}
x={4 pm sqrt{16-56}}/{4}
x={4 pm sqrt{-40}}/{4}
Simplify the radical by looking for perfect square factors of 40.
x={4 pm sqrt{-40}}/{4}
x={4 pm sqrt{-1(4)(10)}}/{4}
x={4 pm 2i sqrt{10}}/{4}
Simplify by canceling the common factor of 2 out of the terms in the numerator and denominator.
x={4 pm 2i sqrt{10}}/{4}
x={2(2) pm 2i sqrt{10}}/{2(2)}
x={2 pm i sqrt{10}}/{2}

The solutions to the quadratic equation are x={2 + i sqrt{10}}/{2} and x={2 - i sqrt{10}}/{2}.

1.4 Quadratic Equations

Solving Quadratic Equations by Factoring: Trinomial a=1

Example: Solve the quadratic equation by factoring.

x^2+x=20

Solution:

The original equation
x^2+x=20
Write the equation with all the terms on one side of the equation and zero on the other side of the equation. x^2+x=20
x^2+x-20=20-20
x^2+x-20=0
Factor the expression on one side. x^2+x-20=0
(x+5)(x-4)=0
Use the zero product property and set each factor equal to zero. (x+5)(x-4)=0
x+5=0 or x-4=0
Solve each equation. x+5=0 or x-4=0
x+5-5=0-5 or x-4+4=0+4
x=-5 or x=4

Check: x=4

x^2+x=20
(4)^2+4=20
16+4=20
20=20

Since the value of 4 makes the equation true, 4 is a solution to the equation.

Check: x=-5

x^2+x=20
(-5)^2+(-5)=20
25-5=20
20=20

Since the value of -5 makes the equation true, -5 is a solution to the equation.

1.4 Quadratic Equations

Solving a Quadratic Equation: The Square Root Method Example 1 of 1

Example: Solve the quadratic equation with completing the square.

x^2-9=0

The original quadratic equation.
x^2-9=0
Rewrite the quadratic equation so that the square and everything that the square applies to are on one side of the equation.  This is called isolating the square. x^2-9=0
x^2-9+9=0+9
x^2=9
Cancel out the square by square rooting both sides. x^2=9
sqrt{x^2}=sqrt{9}
x=pm 3
The remaining equations are already solved.  The solutions to the equation are 3 and -3. x=-3 or x=3
1.2 Applications of Linear Equations, Applications of Linear Equations

Application of Rational Equations: Work Together (one pump stops working)

Problem:  Two pumps were required to pump the water out of a submerged area after a flood.  Pump A, the larger of the two pumps, can pump the water out in 24 hours, whereas it would take pump B 120 hours.  Both pumps were working for the first 8 hours until pump A broke down.  How long did it take pump B to pump the remaining water?

Solution:

Create a table.  List each individual and the time it takes to complete the job.  Also include a row for the pumps working together.  Use a variable to represent the unknown time to complete the job when the pumps are working together.

Time to complete the job
Pump A
  24 hours
Pump B
 120 hours
Together x

Fill in the table with the portion of the job completed in one hour.

If it takes Pump A 24 hours to complete the whole job, one twenty-forth (1/24) of the job will be completed in one hour.

If it takes Pump B 120 hours to complete the whole job, one one hundred twentieth (1/120) of the job will be completed in one hour.

If it takes x number of hours when the pumps are working together 1/x portion of the job will be completed in one hour.

Time to complete the job
  Portion completed in 1 hour
Pump A
 24 hours  1/24
Pump B
 120 hours  1/120
Together x  1/x

From here an equation can be created with the portion of the job completed in one hour.

In one hour, the portion completed by pump A plus the portion completed by pump B should equal the portion when they are working together.

1/24+1/120=1/x

Solve the equation to find x.

 The original equation is a rational equation.
1/24+1/120=1/x
Multiply the least common multiple of the denominators by each term.  The least common multiple of 24, 120 and x is 120x. 
1/24+1/120=1/x
120x(1/24)+120x(1/120)=120x(1/x)
Simplify by canceling the common factors.
 120x(1/24)+120x(1/120)=120x(1/x)
5x+x=120
Solve the remaining linear equation. 5x+x=120
6x=120
{6x}/6={120}/6
x=20

It would take both pumps working together 20 hours to pump out all of the water.

But the two pumps are only working together for 8 hours which means they only get eight-twentieths (8/20) of the job done.  This fraction reduces to two-fifths (2/5).

Three-fifths (3/5) of the water remains and pump B is working alone.

It takes pump B 120 hours to complete the whole job.  It will take 120(3/5) to pump the remaining water out.

120(3/5)= 72 hours.

It takes pump B 72 hours to pump the remaining water.

1.8 Absolute Value Equations and Inequalities

Absolute Value Inequality: Less than or less than or equal to

Example: Solve the inequality.  Express the solution using interval notation.

delim{|}{12-4x}{|}<=3

Solution:

Original absolute value inequality
delim{|}{12-4x}{|}<=3
Since the absolute value is isolated, get rid of the absolute value by writing the equivalent compound inequality.
delim{|}{12-4x}{|}<=3
-3<=12-4x<=3
Solve the compound inequality by isolating the x in the middle. Start by subtracting 12 from each part.
-3<=12-4x<=3
-3-12<=12-12-4x<=3-12
-15<=-4x<=-9
Isolating the x in the middle: Divide by -4 on each part.
Note: Reverse the inequality symbol when dividing by a negative.  Reversing the order of the numbers is equivalent to reversing the signs.
-15<=-4x<=-9
{-9}/{-4}<={-4x}/{-4}<={-15}/{-4}
9/4 <=x <=15/4
Write the interval notation for the inequality.
[9/4,15/4]