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## Quadratics: Standard Form to Vertex Form

Q:  Write the equation of y = -x2 + 2x + 2 in the form of y = a(x-h)2 + k

A:  To do this, we are going to use a strategy called “completing the square” — a fairly complicated algebra 2 concept.

We are currently in the form y = ax2+bx+c and we want to get to y = a(x-h)2 + k

First, let’s establish a, b, and c in our equation: y = -x2 + 2x + 2

a = -1, b = 2, c = 2

I will walk you through the steps on how to do the problem.  At the end, I will provide some explanation behind the concept and the “why”.

Step 1)  Divide the entire equation by “a”

So, divide everything by -1:

y = -x2 + 2x + 2

y/ -1 = (-x2 + 2x + 2) / -1

And simplify to get:

-y = x2 – 2x – 2

Step 2)  “Complete the Square”

This step involves finding what number to add (or subtract) into the equation that will make the x terms factor nicely.  To find this number, we follow a simple pattern:

Take 1/2 of the coefficient in front of the x term and then square it.

The coefficient in front of x is -2.

1/2 * -2 = -1

Square -1 to get +1.

The number we need to “Complete the Square” is +1.

Add this number to both sides of the equation (remember to do it to BOTH SIDES to keep the equation balanced).  Also, add in by the x’s just for ease, like so:

-y + 1 = x2 – 2x + 1 – 2

So, our modified equation is:

-y + 1 = x2 – 2x + 1 – 2

Step 3)  Factor the x terms:

We are now going to factor the part of the equation I’ve highlighted.  The part we factor involves the x’s and the “complete the square” number that we added to the equation:

-y + 1 = x2 – 2x + 1 – 2

Factoring x2 – 2x + 1 gives (x – 1)(x – 1) OR (x – 1)2

So, we now have:

-y + 1 = (x – 1)2 – 2

Step 4) Isolate y

We have finished the hardest part (completing the square)!  Now, we just solve for y and we are done:

-y + 1 = (x – 1)2 – 2

Subtract 1 from both sides:

-y = (x – 1)2 – 3

Multiply everything by -1:

y = -(x – 1)2 + 3

And there it is!  We have taken the original equation: y = -x2 + 2x + 2 and re-written it in a different form: y = -(x – 1)2 + 3

Some logic behind the process:

Completing the square is the process of figuring out “what number” is needed to add (or subtract) in the equation so that it will factor easily into something like (x-h)2.  Once we determine the number needed, we add it to both sides of the equation to maintain balance.  Remember, we aren’t altering the actual equation, we are just changing its appearance.  Once we found the correct number, the equation will factor the way we need it to.

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## Multiplying Polynomials

Q: Multiply -9m(2m2 + 6m – 1)

A:  When adding and subtracting polynomials, you can only combined like terms.  This is not the case with multiplication.  You can multiply unlike terms together.

-9m(2m2 + 6m – 1)

In this problem the whole quantity in the parentheses is being multiplied by -9m.  So, -9m needs to multiply each term:

-9m(2m2 + 6m – 1) = -9m*2m2 + -9m*6m + -9m*-1
Now we need to clean up the expression on the right:

-9m*2m2 + -9m*6m + -9m*-1 ….

-9m*2m2 can be thought of as -9*m*2*m*m = -18m3

(multiply the coefficients together, and multiply the m’s together)

-9m*6m = -54m2

-9m*-1 = 9m

So, final answer cleans up like:

-9m(2m2 + 6m – 1) = -18m3 – 54m2 + 9m

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## Subtracting Polynomials

1:  Subtract: (-4y2-3y+8) – (2y2-6y+2)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

y2 terms get combined with y2 terms
y terms get combined with y terms
constants (lone numbers) get combined with constants

So, let’s use color to highlight the like terms:

(-4y2-3y+8) – (2y2-6y+2)

How many total y2 terms do you have? -4y2 – 2y2 = -6y2

How many total y terms do you have? -3y – -6y = -3y+6y = 3y

How many total constant terms do you have? 8+2 = 8 – 2 = 6

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(-4y2-3y+8) – (2y2-6y+2) = -6y2 + 3y + 6

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A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

x3 terms get added with x3 terms
x2 terms get added with x2 terms
x terms get added with other x terms

So, let’s use color to highlight the like terms:

(3x3+2x2-5x) + (-4x3x2-8x)

How many total x3 terms do you have? 3x3+ -4x3 = -1x3

How many total x2 terms do you have? 2x2+x2 = 1x2

How many total x terms do you have? -5x+ -8x = -13x

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

(3x3+2x2-5x) + (-4x3x2-8x) = -1x3 + 1x2 13x

Since we tend not to write the “1”, you might see the answer displayed as:

x3 + x2 13x

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## Solving logarithmic equations

Q:  Solve
log16x + log4x + log2x = 7

A:  It is easier (though not always necessary) to have all of the logs in the same base to proceed.  Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
logab = logcb / logcwhere c can be any base of your choosing

What base should we go to?  Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log16x + log4x + log2x = 7

log2x / log216 + log2x / log24 + log2x / log22 = 7

Now, we can simplify:

log2x / 4 + log2x / 2 + log2x / 1 = 7

1/4 log2x + 1/2 log2x + log2x = 7

7/4 log2x = 7

log2x = 7 *(4/7)

log2x = 4

(remember, this reads:  the power you put on 2 to get x is 4)

x = 24

x = 16

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## Solving Exponential Equations by Factoring

Q:  Solve for x:
22x + 2x+2 – 12 = 0

A:  The first thing we need to notice (from practice and experience) is that we can re-write this like so using our knowledge / rules of exponents:

(2x)2 + 22*2x – 12 = 0
Now simplify a little:

(2x)2 + 4*2x – 12 = 0

So, look at this in a new light.  What if we substitute each 2x with y?

(2x)2 + 4*2x – 12 = 0   turns into   y2 + 4*y – 12 = 0

(this isn’t necessary, but is helpful for visualization)

We see this is in the form of a quadratic and can be factored:

(y + 6)(y – 2) = 0

So, y = -6 or y = 2

Remember, y was a substitution for 2x. So, we really have:
2x = -6   or   2x = 2

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## Solving Percent Problems using Proportions

Q:  How do you solve percent problems using proportions?

A:  Generally, there are 3 main parts to a percent problem:  the percent, the part and the total.

The set up is the same for all problems:

part/total = percent/100

(remember, percent means per 100, so it is always out of 100)

Example 1:  Finding the percent:

What percent is 75 of 85?

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## Solve the system of equations (conics)

Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

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## Negative Exponent Examples and Basics

Q:  What is a negative exponent?  What does a negative exponent do?

A:  A negative exponent is just notation that can be used to represent a reciprocal.  Let me show you some examples:

x – 1 = 1/x1 or 1/x

x-2 = 1/x2

4 – 1 = 1/41 or 1/4

4 – 2 = 1/42 = 1/16

Those are the basics.  Let’s look at a few more complicated examples of how negative exponents affect an expression:

3x – 4 = 3/x4

4x6y – 3 = 4x6/y3

OR… If the negative exponent is affecting something in a denominator, it will move it to the numerator like so:

1/x – 5 = 1*x5 = x5

6/y – 7 = 6y7

These are just a few basic examples to show you how negative exponents work. Of course things can get more difficult.

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Q:  Solve for x:  (x + 3)(x – 4) < 0

A:

Step 1:  Find the zeroes

Since the quadratic is already factored, this isn’t too tricky.  If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 or x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line:

Now, you have to test each “interval” that is separated by the “zeroes”.  There are three intervals to test.

Interval 1:  The numbers to the left of -3  –> in interval notation this is (-infinity, -3)

Interval 2:  The numbers between -3 and 4  –> in interval notation this is (-3, 4)

Interval 3:  The numbers to the right of 4  –> in interval notation this is (4, infinity)

Step 2:  Test each interval

Pick any number on interval 1 and test it into the original inequality.  I’ll pick -5:

(x + 3)(x – 4) < 0

(-5 + 3)(-5 – 4) < 0

(-2)(-9) < 0

18 < 0      <—  this is false, so numbers on this interval [interval 1] are not part of the solution.

Pick any number on interval 2 and test it into the original inequality.  I’ll pick 0:

(x + 3)(x – 4) < 0

(0 + 3)(0 – 4) < 0

(3)(-4) < 0

-12 < 0      <—  this is true, so numbers on this interval [interval 2] are a part of the solution.

Pick any number on interval 3 and test it into the original inequality.  I’ll pick 5:

(x + 3)(x – 4) < 0

(5 + 3)(5 – 4) < 0

(8)(1) < 0

8 < 0      <—  this is false, so numbers on this interval [interval 3] are not part of the solution.

SO:  The only interval that “worked” was interval 2.  Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4

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Q:  Solve  (a/2)2 – a > 0

Solving Algebraically:

Step 1:  Simplify

We can simplify the equation by squaring the fraction to get:

a2/4 – a > 0

OR in different form:

(1/4) a2 – a > 0

Now, to get rid of the fraction and clean things up, I am going to multiply everything by 4 (since the fraction is 1/4 — you don’t have to do this, just a “cleaning up” step”):

4*(1/4) a2 – 4*a > 4*0

a2 – 4a > 0

Normally, when you solve inequalities, you isolate the variable by moving things around to the left / right side.  When it is a quadratic, you don’t want to do that.  You want all of the numbers and variables on one side and zero on the other side.  We have that, so we are good to go.

Step 2:  Factor

Now, factor the left side:

a2 – 4a > 0

a (a – 4) > 0

Step 3:  Identify the zeroes

As we are used to doing with quadratics, we need to find what values you plug in to make “zeroes”

So, take each factor and set it equal to zero like so:

a = 0 and a – 4 = 0

Solve to get:

a = 0 and a = 4

The zeroes are:  0 and 4.

So, this parabola (quadratic) crosses the x-axis at 0 and 4.  Now, we want to find where the quadratic is greater than 0.

With analysis, we know the quadratic is concave up in shape (a smiley face).  On your paper, draw an x-y graph with a parabola that crosses the x-axis and 0 and 4 and is concave up.  You should get:

So, where is the parabola > 0 (above the x-axis?)

When a < 0 and when a > 4

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## Graphing Lines by Plotting Points

Q:  Graph x + 2y = 4

A:  We are going to graph this by making an x-y table of values.  A typical (though not necessary) thing to do is solve for y first — this just means “get y by itself”

x + 2y = 4

Subtract x from both sides (to try to get y alone)

2y = 4 – x

Now, it’s standard for to list the x first, so we may write this as:

2y = -x + 4

To get y by itself, divide everything by 2:

y = -x/2 + 4/2

This simplifies to:

y = -1/2 x + 2

So, to make a table of values, pick some numbers for x.  It is standard to pick “0, 1, 2” or something similar.  So:

x  |  0  |  1  |  2  |

y  |      |      |      |

Plug in each value for x to find y:

When x = 0, y = -1/2 (0) + 2 = 0 + 2 = 2

When x = 1,  y = -1/2 (1) + 2 = -1/2 + 2 = 1.5

When x = 2, y = -1/2 (2) + 2 = -1 + 2 = 1

Then the table is complete:

x  |  0  |  1    |  2  |

y  |  2  | 1.5  |  1  |

These are just points you need to plot on a graph:  (0, 2)  (1, 1.5)  and (2,1)

Plot the points then connect the dots to make a line!