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QUADRATIC EQUATIONS

Any equation of type ax2 + bx + c = 0 where a, b, and c are constants and a <> 0, is in standard form for a quadratic equation.
Quadratic equations of type ax2 + bx + c = 0 and ax2 + bx = 0 (c is 0) can be factored to solve for x. Examples:
1. Problem: Solve 3x2 + x - 2 = 0 for x.
Solution: Factor.
(3x - 2)(x + 1) = 0
Use the principle of zero products,
which says, if ab = 0, either
a, b, or both must be
equal to zero.
3x - 2 = 0, x + 1 = 0
3x = 2 , x = -1
x = (2/3)
x = -1, (2/3)

2. Problem: Solve 3x2 + 5x = 0 for x.
Solution: Factor.
x(3x + 5) = 0
Use the principle of zero products.
x = 0, 3x + 5 = 0
3x = -5
x = -(5/3)
x = 0, -(5/3)
Quadratic equations of type ax2 + c = 0 can be solved by solving for x. Example:

3. Problem: Solve 3x2 = 6 for x.
Solution: Recognize that the equation is quadratic because it is the same as 3x2 - 6 = 0.
Divide each side by 3.
x2 = 2
Take the square root of each side.
x = SQRT(2), -(SQRT(2))
Many times you will come across quadratic equations that are not easy to factor or solve. In those cases, there is a special formula called the quadratic formula that you can use to solve any quadratic equation.
The solutions of any quadratic equation, ax2 + bx + c = 0 is given by the following formula, called the quadratic formula:

-b ± SQRT(b2 - 4ac)
x = -------------------
2a
Example:

1. Problem: Solve 3x2 + 5x = -1 for x.

Solution: First find the standard form of
the equation and determine a, b,
and c.

3x2 + 5x + 1 = 0
a = 3
-5 ± SQRT(52 - 4(3)(1))
x = -----------------------
2 * 3

b = 5
c = 1

Plug the values you found for a, b, and c into the quadratic formula. Perform any indicated operations.

-5 ± SQRT(25 - 12)
x = ------------------
6

-5 ± SQRT(13)
x = -------------
6
The solutions are as follows:

-5 + SQRT(13) -5 - SQRT(13)
x = -------------, -------------
6 6
Some equations are not quadratic equations, but are in the same form, such as x4 - 9x2 + 8 = 0. To solve equations such as that, you make a substitution, solve for the new variable, and then solve for the original variable. Example:

1. Problem: Solve x4 - 9x2 + 8 = 0 for x.

Solution: Let u = x2. Then substitute u for every x2 in the equation.
u2 - 9u + 8 =0
Factor.
(u - 8)(u - 1) = 0
Utilize the principle of zero products.
u - 8 = 0, u - 1 = 0
u = 8 , u = 1
Now substitute x2 for u and solve the equations.

x2 = 8, x2 = 1
x = ±SQRT(8), x = ± 1
x = ±2(SQRT(2))

x = ±2(SQRT(2)), ±1

Another way to solve systems of equations is by substitution. In this method, you solve on equation for one variable, then you substitute that solution in the other equation, and solve. Example:
1. Problem: Solve the following system:
x + y = 11
3x - y = 5

Solution: Solve the first equation for y (you could solve for x - it doesn't matter).
y = 11 - x
Now, substitute 11 - x for y in the second equation. This gives the equation one variable, which earlier algebra work has taught you how to do.

3x - (11 - x) = 5
3x - 11 + x = 5
4x = 16
x = 4
Now, substitute 4 for x in either equation and solve for y. (We use the first equation below.)

4 + y = 11
y = 7

The solution is the ordered pair,
(4, 7).
The last method, addition, is probably the most complicated, but is necessary when dealing with more complex systems, such as systems with three or more variables. The idea behind the addition method is to replace an equation with a combination of the equations in the system. To obtain such a combination, you multiply each equation by a constant and add. You choose the constants so that the resulting coefficient of one of the variables will be 0. Example:

2. Problem: Solve the following system:
5x + 3y = 7
3x - 5y = -23
Solution: Multiply the second equation by 5 to make the x-coefficient a multiple of 5. (This works because it does not change the equation (see the multiplication property).)

15x - 25y = -115

Next, multiply the first equation by -3 and add it to the second equation. This gets rid of the x-term.

-15x - 9y = -21
15x - 25y = -115
-----------------
- 34y = -136

Now, solve the second equation for y. Then substitute the result into the first equation and solve for x.

-34y = -136
y = 4

5x + 3(4) = 7
5x + 12 = 7
5x = -5
x = -1
The solution is the ordered pair, (-1, 4).


Since you would need a three-dimensional coordinate system to solve systems in three variables, solving graphically is not an option. Substitution would work, but is usually unmanageable. Therefore, we will use the addition method, which is basically the same process as it is with systems in two variables.
1. Problem: Solve the following system:
x + y + z = 4
x - 2y - z = 1
2x - y - 2z = -1

Solution: Start out by multiplying the first equation by -1 and add it to the second equation to eliminate x from the second equation.

-x - y - z = -4
x - 2y - z = 1
----------------
-3y - 2z = -3

Now eliminate x from the third equation by multiplying the first equation by -2 and add it to the third equation.
-2x - 2y - 2z = -8
2x - y - 2z = -1
------------------
-3y - 4z = -9

Next, eliminate y from the third equation by multiplying the second equation by -1 and adding it to the third equation.

3y + 2z = 3
-3y - 4z = -9
--------------
-2z = -6

Solve the third equation for z.

-2z = -6
z = 3

Substitute 3 for z in the second equation and solve for y.

-3y - 2z = -3
-3y - 2(3) = -3
-3y - 6 = -3
-3y = 3
y = -1

Lastly, substitute -1 for y and 3 for z in the first equation and solve for x.

x + (-1) + 3 = 4
x + 2 = 4
x = 2

The answer is (2, -1, 3).


The easiest way to solve systems of inequalities is to solve them by graphing. Therefore, it is best if you know how to graph inequalities in two variables (5x - 4y < 13, for example). Example:
1. Problem: Graph y < x.

Solution: First graph the equation y = x. However, the line must be drawn dashed because the less than sign tells us the line is not included in the solution. Next, test a point that is located above the line and one that is below the line. Any point you pick above the line, such as (0, 2), y is greater than x, so points above the line are not in-cluded in the solution. Points below the line, such as (3, -3) have a y value that is less than the x value, so all points below the line are included in the solution.



To solve a system or conjunction of inequalities, it is easiest to graph each of the inequalities and then find their intersection. Example:
1. Problem: Graph the following system:
2x + y >= 2
4x + 3y <= 12
(1/2) <= x <= 2
y >= 0

Solution: See the figure below.

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