View source for Solving Equations, Absolute Value, and Quadratic Equations

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==Introduction==
Equations are how we represent problems we want to solve. Questions such as: how many months will it take me to payback my friend $500 if I pay him $25 a month or
how high will a baseball go if launched at 35 ft/sec at an angle of 45 degrees, are examples of questions that can be modeled with equations. In this section we
will learn about certain types of equations, and the methods we use to solve them. The main method for solving equations is to transform the initial equation into
an equivalent, but simpler, equation.

==Operations that result in equivalent equations==
We start with the five operations we can use that result in equivalent equations

1. Interchanging the two sides of the equation:
	Replacing 2 = 3x + 4 with 3x + 4 = 2
	
2. Simplifying either side of an equation:
	Replacing 3(x + 2) = 7 with 3x + 6 = 7
	
3. Add/Subtract the same expression from both sides of the equation:
	Replacing 5x + 7 = 8 with 5x = 1
	
4. Multiply/Divide both sides of the equation by the same expression:
	Replacing <math> \frac{5}{x + 2} = \frac{4}{x - 1}</math> with <math> 5(x - 1) = 4(x + 2)</math>

5. If one side of the equation is 0, we can factor the other side and set each factor equal to 0, using the zero product property (if the product of some numbers is zero, at least one of them is zero).
	Example: <math> x^2 - 3x = 0</math> replaced with <math> x(x - 3) = 0</math> and obtaining two new equations x = 0 and x - 3 = 0.
	
==Examples==

Solve the following equation: <math> x^2 = -3x </math>

Solution:

::<math> \begin{array}{rcll}
x^2 & = & -3x & \text{Initial Problem}\\
x^2 + 3x & = & 0 & \text{Adding 3x to both sides}\\
x(x + 3)& = & 0 & \text{Using operation 5, and factoring the left hand side}\\
x = 0 &\text{or}& x + 3 = 0  &\text{The second step of operation 5, setting each factor equal to zero}\\
x = 0 &\text{or}& x = -3 &\text{Adding -3 to both sides of the second equation}
\end{array} </math>

==Absolute Value==
The absolute value function takes the value of a number, regardless of whether it is positive or negative. Some problems that the absolute value is useful for modeling include an object bouncing on the ground.

'''For example,''' <math> \left| -3 \right| = \left|3\right| = 3 </math>, since we do not care about the negative sign.

'''Example:''' Solve <math> \left| x - 3 \right| = 2.</math>

Solution: Since the absolute value does not care about whether a number is positive or negative, if <math> \left| x - 3\right| = 2 </math> then <math>
x - 3 = 2</math> or <math> x - 3 = -2 </math>. Then we can solve each equation individually to find that x = 1 or 5.

==Quadratic Equations==
Quadratic equations are equations that when rearranged using the 5 operations can yield a polynomial of degree 2 on one side of the equation and 0 on the other. There are 3 commonly taught methods for solving
quadratic equations. These methods are factoring, completing the square, and the quadratic formula. These functions are most commonly used to model projectiles, disregarding air resistance.

When using the factoring method one attempts to write a degree 2 polynomial as the product of two degree 1 polynomials. This method works because if we multiply two degree one polynomials together we find
<math> (x + a)(x + b) = x^2 + (a + b)x + ab.</math>

'''For example:''' When factoring <math> x^2 + 5x + 6</math>, we want to find two number a, b whose product is 6 and sum to 5. The values of a and b we are looking for are 2 and 3.

The method of completing the square centers around using the five operations to replace the original equation with an equivalent equation of the form <math> (x + a)^2 = b</math> by taking advantage of the fact
that <math> (x + a)^2 = x^2 + 2ax + a^2</math>

'''Example:''' Solve <math> x^2 + 8x = 21</math>

Solution: Since we want to find an equation of the form <math>x^2 + 2ax + a^2 = b </math>. Since we already have 2a, 8, we find that a = 4. Thus adding 16 to both sides we have an equivalent quadratic equation
of the desired form. Now we have <math>x^2 + 8x + 16 = 21 + 16 = 37 </math>. By the note above the left hand side factors as <math> (x + 4)^2 </math>. Thus, <math> (x+4)^2 = 37</math>, 
<math> x + 4 = \sqrt{37}</math>, and <math> x + 4 = -\sqrt{37}</math>. Finally, <math> x = -4 + \sqrt{37}, -4 - \sqrt{37}</math>.


The final method is the quadratic formula. Given a quadratic equation of the form <math> ax^2 + bx + c = 0</math> the quadratic formula gives the solutions as <math> \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}</math>.

'''Example:''' Find the solutions of <math>2x^2 - 4x + 1 = 0</math>

Solution: By the quadratic formula the solutions are <math> \frac{4 \pm \sqrt{16 - 8}}{2*2} = \frac{4 \pm \sqrt{8}}{4} = \frac{2 \pm \sqrt{2}}{2}</math>

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