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.

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