Week of January 29,2018

Algebra 2

Complex Numbers

 
A Complex Number

A Complex Number is a combination of a
Real Number and an Imaginary Number

Real Numbers are numbers like:

1

12.38

−0.8625

3/4

√2

1998

Nearly any number you can think of is a Real Number!

Imaginary Numbers when squared give a negative result.

Normally this doesn't happen, because:

• when we square a positive number we get a positive result, and
• when we square a negative number we also get a positive result (because a negative times a negative gives a positive), for example −2 × −2 = +4

But just imagine such numbers exist, because we will need them.

The "unit" imaginary number (like 1 for Real Numbers) is i, which is the square root of −1

Because when we square i we get −1

i² = −1

Examples of Imaginary Numbers:

3i

1.04i

−2.8i

3i/4

(√2)i

1998i

And we keep that little "i" there to remind us we need to multiply by √−1

Complex Numbers

A Complex Number is a combination of a Real Number and an Imaginary Number:

Examples:

1 + i

39 + 3i

0.8 − 2.2i

−2 + πi

√2 + i/2

Can a Number be a Combination of Two Numbers?

Can we make up a number from two other numbers? Sure we can!

We do it with fractions all the time. The fraction ³/₈ is a number made up of a 3 and an 8. We know it means "3 of 8 equal parts".

Well, a Complex Number is just two numbers added together (a Real and an Imaginary Number).

Either Part Can Be Zero

So, a Complex Number has a real part and an imaginary part.

But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers.

Complex Number	Real Part	Imaginary Part
3 + 2i	3	2
5	5	0
−6i	0	−6

Complicated?

Complex does not mean complicated.

It means the two types of numbers, real and imaginary, together form a complex, just like a building complex (buildings joined together).

A Visual Explanation

You know how the number line goes left-right?

Well let's have the imaginary numbers go up-down:

And we get the Complex Plane

A complex number can now be shown as a point:

The complex number 3 + 4i

Adding

To add two complex numbers we add each part separately:

(a+bi) + (c+di) = (a+c) + (b+d)i

Example: add the complex numbers 3 + 2i and 1 + 7i

• add the real numbers, and
• add the imaginary numbers:

(3 + 2i) + (1 + 7i)
= 3 + 1 + (2 + 7)i
= 4 + 9i

Let's try another:

Example: add the complex numbers 3 + 5i and 4 − 3i

(3 + 5i) + (4 − 3i)
= 3 + 4 + (5 − 3)i
= 7 + 2i

On the complex plane it is:

Multiplying

To multiply complex numbers:

Each part of the first complex number gets multiplied by
each part of the second complex number

Just use "FOIL", which stands for "Firsts, Outers, Inners, Lasts" (see Binomial Multiplication for more details):

	Firsts: a × c Outers: a × di Inners: bi × c Lasts: bi × di
(a+bi)(c+di) = ac + adi + bci + bdi2

Like this:

Example: (3 + 2i)(1 + 7i)

(3 + 2i)(1 + 7i)= 3×1 + 3×7i + 2i×1+ 2i×7i

 = 3 + 21i + 2i + 14i²

 = 3 + 21i + 2i − 14  (because i² = −1)

 = −11 + 23i

And this:

Example: (1 + i)²

(1 + i)(1 + i)= 1×1 + 1×i + 1×i + i²

 = 1 + 2i − 1  (because i² = −1)

 = 0 + 2i

But There is a Quicker Way!

Use this rule:

(a+bi)(c+di) = (ac−bd) + (ad+bc)i

Example: (3 + 2i)(1 + 7i) = (3×1 − 2×7) + (3×7 + 2×1)i = −11 + 23i

Why Does That Rule Work?

It is just the "FOIL" method after a little work:

(a+bi)(c+di) =ac + adi + bci + bdi²  FOIL method

 =ac + adi + bci − bd  (because i² = −1)

 =(ac − bd) + (ad + bc)i  (gathering like terms)

And there we have the (ac − bd) + (ad + bc)i  pattern.

This rule is certainly faster, but if you forget it, just remember the FOIL method.

Let us try i²

Just for fun, let's use the method to calculate i²

Example: i²

We can write i with a real and imaginary part as 0 + i

i² = (0 + i)²= (0 + i)(0 + i)

 = (0×0 − 1×1) + (0×1 + 1×0)i

 = −1 + 0i

 = −1

And that agrees nicely with the definition that i² = −1

So it all works wonderfully!

Learn more at Complex Number Multiplication.

Conjugates

We will need to know about conjugates in a minute!

A conjugate is where we change the sign in the middle like this:

A conjugate is often written with a bar over it:

Example:

5 − 3i   =   5 + 3i

Dividing

The conjugate is used to help complex division.

The trick is to multiply both top and bottom by the conjugate of the bottom.

Example: Do this Division:

2 + 3i4 − 5i

Multiply top and bottom by the conjugate of 4 − 5i :

2 + 3i4 − 5i×4 + 5i4 + 5i  =  8 + 10i + 12i + 15i²16 + 20i − 20i − 25i²

Now remember that i² = −1, so:

=  8 + 10i + 12i − 1516 + 20i − 20i + 25

Add Like Terms (and notice how on the bottom 20i − 20i cancels out!):

=  −7 + 22i41

Lastly we should put the answer back into a + bi form:

=  −7 41 + 2241i

DONE!

Yes, there is a bit of calculation to do. But it can be done.

Multiplying By the Conjugate

There is a faster way though.

In the previous example, what happened on the bottom was interesting:

(4 − 5i)(4 + 5i) = 16 + 20i − 20i − 25i²

The middle terms (20i − 20i) cancel out! Also i² = −1 so we end up with this:

(4 − 5i)(4 + 5i) = 4² + 5²

Which is really quite a simple result. The general rule is:

(a + bi)(a − bi) = a² + b²

We can use that to save us time when do division, like this:

Example: Let's try this again

2 + 3i4 − 5i

Multiply top and bottom by the conjugate of 4 − 5i :

2 + 3i4 − 5i×4 + 5i4 + 5i  =  8 + 10i + 12i + 15i²16 + 25

=  −7 + 22i41

And then back into a + bi form:

=  −7 41 + 2241i

DONE!

https://www.mathsisfun.com/numbers/complex-numbers.html




Algebra 1

Solving Inequalities

Sometimes we need to solve Inequalities like these:

Symbol	Words	Example
>	greater than	x + 3 > 2
<	less than	7x < 28
≥	greater than or equal to	5 ≥ x - 1
≤	less than or equal to	2y + 1 ≤ 7

Solving

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

Something like:		x < 5
or:		y ≥ 11

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

Simplify:

x > 10

Solved!

How to Solve

Solving inequalities is very like solving equations ... we do most of the same things ...

... but we must also pay attention to the direction of the inequality.

Direction: Which way the arrow "points"

Some things can change the direction!

< becomes >

> becomes <

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

• Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

3x < 10

But these things do change the direction of the inequality ("<" becomes ">" for example):

Multiply (or divide) both sides by a negative number
Swapping left and right hand sides

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality:

12 > 2y+7

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra), like this:

Solve: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 - 3 < 7 - 3    

x < 4

And that is our solution: x < 4
In other words, x can be any value less than 4.

What did we do?

We went from this: To this:				x+3 < 7 x < 4

And that works well for adding and subtracting, because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5
If we subtract 5 from both sides, we get:

12 - 5 < x + 5 - 5    

7 < x

That is a solution!
But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

x > 7

Do you see how the inequality sign still "points at" the smaller value (7) ?
And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number:

Solve: 3y < 15

If we divide both sides by 3 we get:

3y/3 < 15/3

y < 5

And that is our solution: y < 5

Negative Values

When we multiply or divide by a negative number  we must reverse the inequality.

Why?

Well, just look at the number line!

For example, from 3 to 7 is an increase,
but from -3 to -7 is a decrease.

-7 < -3	7 > 3

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Solve: -2y < -8

Let us divide both sides by -2 ... and reverse the inequality!

-2y < -8

-2y/-2 > -8/-2

y > 4

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Solve: bx < 3b

It seems easy just to divide both sides by b, which gives us:

x < 3

... but wait ... if b is negative we need to reverse the inequality like this:

x > 3

But we don't know if b is positive or negative, so we can't answer this one!

To help you understand, imagine replacing b with 1 or -1 in the example of bx < 3b:

• if b is 1, then the answer is x < 3
• but if b is -1, then we are solving -x < -3, and the answer is x > 3

The answer could be x < 3 or x > 3 and we can't choose because we don't know b.

So:

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Solve: (x-3)/2 < -5

First, let us clear out the "/2" by multiplying both sides by 2.
Because we are multiplying by a positive number, the inequalities will not change.

(x-3)/2 ×2 < -5 ×2  

(x-3) < -10

Now add 3 to both sides:

x-3 + 3 < -10 + 3    

x < -7

And that is our solution: x < -7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Solve:

-2 < (6-2x)/3 < 4

First, let us clear out the "/3" by multiplying each part by 3:
Because we are multiplying by a positive number, the inequalities will not change.

-6 < 6-2x < 12

Now subtract 6 from each part:

-12 < -2x < 6

Now multiply each part by -(1/2).
Because we are multiplying by a negative number, the inequalities change direction.

6 > x > -3

And that is the solution!
But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

-3 < x < 6

Summary

• Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.
• But these things will change direction of the inequality:
  • Multiplying or dividing both sides by a negative number
  • Swapping left and right hand sides
• Don't multiply or divide by a variable (unless you know it is always positive or always negative)

https://www.mathsisfun.com/algebra/inequality-solving.html

Week of January 22, 2018

Algebra 2

The Quadratic Formula: Solutions and the Discriminant

• Solve x (x – 2) = 4. Round your answer to two decimal places.

I not only cannot apply the Quadratic Formula at this point, I cannot factor either. Why? Because this equation is yet in the correct form.

And I certainly can not claim, with a straight face, that "x = 4, x – 2 = 4", because this is not how "solving by factoring" works.

No matter which solution method I intend to use — whether I'm factoring or using the Quadratic Formula to find my answers — I must first rearrange the equation into the form "(quadratic) = 0".

 

MathHelp.com

The Quadratic Formula

The first thing I'll do here is multiply through on the left-hand side, and then I'll move the 4 over from the right-hand side to the left-hand side:

x (x – 2) = 4

x2 – 2x = 4

x2 – 2x – 4 = 0

Since there are no factors of (1)(–4) = –4 that add up to –2, then this quadratic does not factor. (In other words, there is no possible way that the faux-factoring solution of "x = 4, x – 2 = 4" could be even slightly correct.)

So factoring won't work, but I can use the Quadratic Formula; in this case, I'll be plugging in the values a= 1, b = –2, and c = –4:

$x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)\,}}{2(1)}$

$= \dfrac{2 \pm \sqrt{4 + 16\,}}{2} = \dfrac{2 \pm \sqrt{20\,}}{2}$

$= \dfrac{2 \pm \sqrt{4\,}\sqrt{5\,}}{2} = \dfrac{2 \pm 2\, \sqrt{5\,}}{2}$

$= \dfrac{2\,\left(1 \pm \sqrt{5\,}\right)}{2(1)} = 1 \pm \sqrt{5\,}$

$\approx 1.236068,\, 3.236968$

Then the answer is:

x = –1.24, x = 3.24, rounded to two decimal places.

---

For reference, here's what the graph of the associated quadratic, y = x2 – 2x – 4, looks like:

As you can see, the solutions from the Quadratic Formula match up with the x-intercepts. The locations where the graph crosses the x-axis give the values that solve the original equation.

There is another connection between the solutions from the Quadratic Formula and the graph of the parabola: you can tell how many x-intercepts you're going to have from the value inside the square root. The argument (that is, the contents) of the square root, being the expression b2 – 4ac, is called the "discriminant" because, by using its value, you can "discriminate" between (that is, be able to tell the difference between) the various solution types.

In this case, the value of the discriminant b2 – 4ac was 20; in particular, the value was not zero and was not negative. Because the value was not negative, the equation was going to have at least one (real-valued) solution; because the value was not zero, the two solutions were going to be distinct (that is, they were going to be different from each other).

---

• Solve 9x2 + 12x + 4 = 0. Leave your answer in exact form.

Using a = 9, b = 12, and c = 4, the Quadratic Formula gives me:

$x = \dfrac{-(12) \pm \sqrt{(12)^2 - 4(9)(4)\,}}{2(9)}$

$= \dfrac{-12 \pm \sqrt{144 - 144\,}}{18}$

$= \dfrac{-12 \pm \sqrt{0\,}}{18} = \dfrac{-12 \pm 0}{18}$

$= \dfrac{-12}{18} = \dfrac{-2}{3} = -\dfrac{2}{3}$

Then the answer is:

$\mathbf{\color{purple}{\mathit{x} = -\dfrac{2}{3}}}$

---

In the first example on this page, I had gotten two solutions because the value of the discriminant (that is, the value inside the square root) was non-zero and positive. As a result, the "plus-minus" part of the Formula gave me two distinct values; one for the "plus" part of the numerator and another for the "minus" part. In this case, though, the square root reduced to zero, so the plus-minus didn't count for anything.

This sort of solution, where you get only one value because "plus or minus zero" didn't change anything, is called a "repeated" root, because x is equal to $- \frac{2}{3}$, but it's equal to this value kind-of twice: $- \frac{2}{3} + 0$ and $- \frac{2}{3} - 0$.

http://www.purplemath.com/modules/quadform2.htm

Algebra 1

SIMPLIFYING RADICALS

To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how.

Here are the steps required for Simplifying Radicals:

Step 1:	Find the prime factorization of the number inside the radical. Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Click on the link to see some examples of Prime Factorization. Also factor any variables inside the radical.
Step 2:	Determine the index of the radical. The index tells you how many of a kind you need to put together to be able to move that number or variable from inside the radical to outside the radical. For example, if the index is 2 (a square root), then you need two of a kind to move from inside the radical to outside the radical. If the index is 3 (a cube root), then you need three of a kind to move from inside the radical to outside the radical.
Step 3:	Move each group of numbers or variables from inside the radical to outside the radical. If there are nor enough numbers or variables to make a group of two, three, or whatever is needed, then leave those numbers or variables inside the radical. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group.
Step 4:	Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Multiply all numbers and variables outside the radical together.

Example 1 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical.
Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and 3’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Example 2 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is three because it is a cube root, which means we need three of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 3’s, x’s (two groups), and y’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems
Example 3 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is five because it is a fifth root, which means we need five of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 3’s, x’s, and y’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems

Example 4 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 2’s, 3’s, x’s (two groups), and y’s (four groups) moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems
Example 5 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is four because it is a fourth root, which means we need four of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and y ’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems

Example 6 – Simplify: 

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is three because it is a cube root, which means we need three of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 2’s, 3’s, x’s, and y’s (two groups) moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

http://www.mesacc.edu/~scotz47781/mat120/notes/radicals/simplify/simplifying.html