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# algebra 2 concept

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## steven buhrow

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#### Transcript of algebra 2 concept

Algebra 2
Part 1 Solving Equations and Inequalities Graphs and Functions Systems of Equations
and Inequalities
Polynomials and Factoring Powers and Roots Complex Numbers Quadratics
Coordinate Geometry
Matrices
Combining Like
Terms Multiplication of
Polynomials Factoring Addition and Multiplication Fractions or Decimals Inequalities Dimensions Graphing Lines Finding Slopes of Lines Functions Solving Systems of Equations Graphically Systems of Inequalities Square Roots
Multiplication and
Simplification Division and Simplification
Operations Rationalizing the Denominator Imaginary Numbers Complex Numbers Equations with
Form Parabolas Distance and Midpoint Formulas Circles Ellipses Hyperbolas Systems of Equations To solve an equation, you isolate the variable you
are solving for. The Addition Principle says that when
a = b, a + c = b + c for any number c. Solve: x + 6 = -15

to each side of the equation.

x + 6 - 6 = -15 - 6

The variable is now isolated.

x = -21
Multiplication
Along the same lines, the Multiplication Principle says that if a = b and c is any number, a * c = b * c. This principle is also used to help isolate the variable you are asked to solve for. Solve: 4x = 9

Solution: Using the Multiplication Principle,
multiply each side of the equation
by (1/4).

(1/4)4x = (1/4)9

The variable is now isolated.

x = (9/4)
2x2 rule 3x3 rule Also, be aware of problems where
you might need to use both of these
principles together! Solve: 3x - 4 = 13

Solution: Use the Addition Principle to

3x - 4 + 4 = 13 + 4

After simplifying, 3x = 17.

Use the Multiplication Principle
to multiply each side by (1/3).

(1/3)3x = (1/3)17

After simplification, the variable
is isolated.

x = (17/3)
When an equation contains fractions or decimals,
it usually makes it easier to solve them when the
fractions or decimals aren't there. The Multiplication
Principle is used to do this. Solve: (3/4)x + (1/2) = (3/2)

Solution: Multiply both sides by the LCM
of the denominators, 4 in
this case.

Use the Distributive Law,
which says a(b + c) = ab + ac
to make the equation easier to
deal with.

(4 * (3/4)x) + (4 * (1/2)) = 4(3/2)

After simplification, you get an
equation with no fractions!

3x + 2 = 6

(It's left up to you to solve
for x.)
If a and b are real numbers, and ab = 0, either a, b, or both equal 0. This principle, called The Principle of Zero Products, is useful when you have an equation to solve that has two instances of a variable, such as (x + 3)(x - 2) = 0. Solve: 7x(4x + 2) = 0

Solution: Using the Principle of
Zero Products,

7x = 0 and 4x + 2 = 0

Solve each equation for x.

x = 0 and x = -(1/2)

The solutions are 0 and -(1/2). Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true.
On many occasions, you will be asked to show the
solution to an inequality by graphing it on a number line. Many times you will have a statement such as x > 5 that needs to be graphed. Because this is not an equation, it does not need to be graphed on the coordinate plane. A number line does the job just fine!

Some things to know about graphin inequalites:
1. An open circle is placed on the number line to show that the number denoted at the circle is not included in the solution set.

2. A circle that is filled in is placed on the number line to show that the number denoted at the circle is included in the solution set. Graph: x < 4

Solution: The problem asks you to graph all numbers
that are less than 4.

Example Graph linerar system Conjunctions or
Complex Inequalities The word conjunction means there are two conditions in a statement that must be met. Therefore, a mathematical statement such as the following is a conjunction: 5 < x < 10. There are two things to remember when dealing with conjunctions. They are outlined below.

1. The greater than or less than signs will always be pointing in the same direction (i.e., you will never see the following: 7 > x < 2).

2. Look out for statements that cannot be true, such as the following: 10 < x < 5 Graph: -2 < x <= 4

Solution: The problem, which is a conjunction, asks for a graph of all the numbers between -2 and 4.
Be sure to note that -2 is not included in the solution while 4 is. Inverse Matrix 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. 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 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 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
b = 5
c = 1

Plug the values you found for
a, b, and c into the

-5 ± SQRT(52 - 4(3)(1))
x = -----------------------
2 * 3

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 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. 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)
Quadratic equations of type ax2 + c = 0 can be solved by solving for x. Problem: Solve 3x2 = 6 for x.

Solution: Recognize that the equation is
same as 3x2 - 6 = 0.

Divide each side by 3.

x2 = 2

Take the square root of each side.

x = SQRT(2), -(SQRT(2)) By definition, every real number has two square roots. For example, 64's two square roots are 8 and -8 because 82 = 64 and (-8)2 = 64. However, the principal square root of a real number is its nonnegative square root (8 would be the principal square root of 64). Problem: SQRT(25)

Solution: 5
That is the answer because 5 * 5 = 25.
One special situation with square roots occurs when the number inside the square root is squared. There is a special theorem that deals with this: SQRT(a2) = |a|. Problem: SQRT((-16)2)

Solution: By the theorem above, the answer
is |-16|. The absolute
value of -16 is 16. adding Multiplying For any nonnegative real numbers, a and b, SQRT(a) * SQRT(b) = SQRT(ab). For example, the 4th root of a times the 4th root of b equals the 4th root of ab. Multiply: SQRT(x + 2) * SQRT(x - 2)

Solution: Use the theorem above to put both

SQRT((x + 2)(x - 2))

Multiply the binomials under the

SQRT(x2 - 4) Reversing the theorem stated above gives us a way to factor radical expressions, thereby simplifying them. Problem: Simplify SQRT(20)

Solution: Factor the radicand as a
product of prime factors.

SQRT(2 * 2 * 5)

There are two instances of 2,
so by the definition of a square
root, you can take 2 out
from under the radicand. That gives

2(SQRT(5)) Taking the square root of fractions and dividing a radicals Which is: When you have a problem like SQRT(27/y2), don't be scared of the fraction. Just use the Roots of Fractions theorem, which says that nRT(a/b) = (nRT(a))/(nRT(b)). Simplify: CBRT(27/125)

Solution: Use the Roots of Fractions
theorem to rewrite the problem.

CBRT(27)
---------
CBRT(125)

Take the cube root of both
the numerator and denominator

(3/5)

Reverse the Roots of Fractions theorem when you are asked to divide a radical by a radical of the same index when it can be simplified. Simplify: (SQRT(80))/(SQRT(5))

Solution: Use the converse of the Roots
of Fractions theorem and rewrite

SQRT(80/5)

80/5 simplifies to 16.

SQRT(16) = 4 You add and subtract radicals the same way you would with polynomials, by combining like terms. You have to look out for terms that do not look alike,
but could be if factored. Problem: 6(SQRT(7)) + 4(SQRT(7))

Solution: Both terms are alike (like

10(SQRT(7)) Problem: 3(SQRT(8)) - 5(SQRT(2))

Solution: Factor 8.
3(SQRT(4 * 2)) - 5(SQRT(2))

Factor SQRT(4 * 2) into two

3(SQRT(4))(SQRT(2)) - 5(SQRT(2))

Take the square root of 4.

3 * 2(SQRT(2)) - 5(SQRT(2))
6(SQRT(2)) - 5(SQRT(2))

Combine like terms.

SQRT(2) To multiply radical expressions that have factors which contain more than one term, use the same procedure you would when multiplying polynomials. Problem: CBRT(y) * (CBRT(y2) + CBRT(2))

Solution: Use the distributive law of
multiplication, which says that
a(b + c) = ab + ac to
multiply the expression out.

CBRT(y) * CBRT(y2) + CBRT(y) * CBRT(2)

CBRT(y3) + CBRT(2y)

Take the cube root of
y3.

y + CBRT(2y) To add matrices, we add the corresponding members. The matrices have to have the same dimensions When dealing with radicals and fractions, you will, on many occasions, get an answer with a radical in the denominator. Usually, an answer is not considered simplified until there are no radicals in the denominator. The process of removing radicals from the denominator is called rationalizing the denominator. An important thing to remember when rationalizing denominators is that anything divided by itself is 1. For example, 67/67 is the same as 1. Problem: Rationalize the denominator.

4 + SQRT(2)
-----------
5 - SQRT(2)

Solution: Multiply by 1 (make sure
the fraction you choose to
use as one will make the
denominator a perfect
square — the conjugate is
usually a good number).

4 + SQRT(2) 5 + SQRT(2)
----------- * -----------
5 - SQRT(2) 5 + SQRT(2)

Multiply the problem as you
would multiply any fractions.
Also, the FOIL method of
multiplying binomials will
come in handy.

20 + 4(SQRT(2)) + 5(SQRT(2)) + (SQRT(2))2
-----------------------------------------
25 + 5(SQRT(2)) - 5(SQRT(2)) - (SQRT(2))2

Perform any indicated operations.

20 + 4(SQRT(2)) + 5(SQRT(2)) + 2
--------------------------------
25 + 5(SQRT(2)) - 5(SQRT(2)) - 2

Perform any indicated operations,
and combine like terms, if you
can.

22 + 9(SQRT(2))
---------------
23 For example, you can multiply a matrix by another matrix or by a number. When you multiply a matrix by a number, multiply each member of the matrix by the number. To multiply a matrix by a matrix, the first matrix has to have the same number of columns as the rows in the second matrix. When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
[(y2 - y1),/(x2 - x1)]. To find the slope, you pick any two points on the line and find the change in y, and then divide it by the change in x When dealing with lines and points, it is very important to be able to find out how long a line segment is or to find a midpoint. Midpoint Formula: [((x1 + x2)/2), ((y1 + y2)/2)]. The distance formula says that the distance d between any two points with coordinates (x1, y1) and (x2, y2) is given by the following equation: d = SQRT[(x2 - x1)2 + (y2 - y1)2]. Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin. When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2 Problem: Find the center and radius of
(x - 2)2 + (y + 3)2 = 16.
Then graph the circle.

Solution: Rewrite the equation in standard form.

(x - 2)2 + [y - (-3)]2 = 42

The center is (2, -3) and the radius is 4.
The graph is easy to draw, especially
if you use a compass.

The figure below is the graph of the solution. Ellipses, or ovals, when centered at the origin, have an equation (standard form) of (x2/a2) + (y2/b2) = 1. When the center of the ellipse is at (h, k), the equation (in standard form) is as follows:

(x - h)2 (y - k)2
-------- + -------- = 1
a2 b2 Problem: Graph x2 + 16y2 = 16.

Solution: Multiply both sides by 1/16
to put the equation in
standard form.

x2 y2
-- + -- = 1
16 1

a = 4 and b = 1. The
vertices are at (±4, 0) and
(0, ±1). (The points are
on the axes because the equation
tells us the center is at the origin,
so the vertices have to be on the
axes.)

Connect the vertices to form an oval,
and you are done!

The figure below is the graph of the ellipse. Absolute Value The equation of a hyperbola (in standard form) centered at the origin is as follows:

x2 y2
-- - -- = 1
a2 b2 Problem: Graph 9x2 - 16y2 = 144.

Solution: First, multiply each side of the
equation by 1/144 to put it
in standard form.

x2 y2
-- - -- = 1
16 9

We now know that a = 4 and
b = 3. The vertices are at
(±4, 0). (Since we
know the center is at the origin,
we know the vertices are on the
x axis.)

The easiest way to graph a hyperbola
is to draw a rectangle using the
vertices and b, which is on
the y-axis.

Draw the asymptotes through opposite
corners of the rectangle.

Then draw the hyperbola.
The figure below is the graph
of 9x2 - 16y2 = 144. The easiest way to solve systems of equations that include circles, ellipses, or hyperbolas, is graphically. Because of the shapes (circles, ellipses, etc.), there can be more than one solution. Problem: Solve the following system of
equations:

x2 + y2 = 25
3x - 4y = 0

Solution: Graph both equations on the same
coordinate plane. The points of
intersection have to satisfy both
equations, so be sure to check the
solutions. Both intersections do
check.

The figure below shows the solution. When terms of a polynomial have the same variables
raised to the same powers, the terms are called similar,
or like terms. Like terms can be combined to make the
polynomial easier to deal with. Problem: Combine like terms in the following equation: 3x2 - 4y + 2x2.

Solution: Rearrange the terms so it is easier
to deal with.

3x2 + 2x2 - 4y

Combine the like terms.

5x2 - 4y Probably the most important kind of polynomial multiplication that you can learn is the multiplication of binomials (polynomials with two terms). An easy way to remember how to multiply binomials is the FOIL method, which stands for first, outside, inside, last. Example :Problem: Multiply (3xy + 2x)(x^2 + 2xy^2).

Solution: Multiply the first terms of each bi-
nomial. (F)

3xy * x2 = 3x3y

Multiply the outside terms of each bi-
nomial. (O)

3xy * 2xy2 = 6x2y3

Multiply the inside terms of each bi-
nomial. (I)

2x * x2 = 2x3

Multiply the last terms of each bi-
nomial. (L)

2x * 2xy2 = 4x2y2

You now have a polynomial with four terms.
Combine like terms if you can

3x3y + 6x2y3 + 2x3 + 4x2y2 Factoring is the reverse of multiplication. When factoring, look for common factors. Problem: Factor out of a common factor of
4y2 - 8.

Solution: 4 is a common factor of
both terms, so pull it out and write
each term as a product of factors.

4y2 - (4)2

Rewrite using the distributive law of
multiplication, which says that
a(b + c) = ab + ac.

4(y2 - 2) Trinomials and binomials are the most common
polynomials, but you will sometimes see polynomials
with more than three terms. Sometimes, when you are
dealing with polynomials with four or more terms, you
can group the terms in such a way that common factors
can be found. Problem: Factor 4x2 - 3x + 20x - 15.

Solution: Rearrange the terms so common
factors can be more easily found.

4x2 + 20x - 3x - 15

The first two terms have a common factor
in 4x. The last two terms have a
common factor in 3. Factor those
terms out.

4x(x + 5) - 3(x + 5)

Now you have a binomial. Each term
has a factor of (x + 5). Factor
that out for the final answer.

(x + 5)(4x - 3) In the set of real numbers, negative numbers do not have square roots. A new kind of number, called imaginary was invented so that negative numbers would have a square root. These numbers start with the number i, which equals the square root of -1, or i2 = -1. All imaginary numbers consist of two parts, the real part, b, and the imaginary part, i. Simplify: SQRT(-5)

Solution: Write -5 as a product of
prime factors.

SQRT(-1 * 5)

Write as separate square roots.

(SQRT(-1))(SQRT(5))

By definition, i = SQRT(-1),
(SQRT(5))i. (SQRT(5)
is the real part, or b.) A complete number system, one that includes both real and imaginary numbers, was devised. Numbers in this set are called complex numbers. Complex numbers consist of all sums a + bi where a and b are real numbers and i is imaginary. Real numbers fit into the complex number system because a = a + 0i.
i behaves as any variable would. Problem: 7i + 9i

Solution: Combine like terms.

16i
Multiplication is done as if the imaginary parts of complex numbers were just another term. Always remember that i2 = -1. Problem: 3i * 4i

Solution: 12i2

Remember that i2 equals

12(-1)
-12 Now that negative numbers have square roots,
equations such as x2 + 1 = 0 have solutions.
They can also be factored! example Problem: Solve for x: x2 + 1 = 0

Solution: Subtract 1 from each side.

x2 = -1

Take the square root of each side.
Remember that i = SQRT(-1).

x = i, -i

2. Problem: Show that (x + i)(x - i) is
a factorization of x2 + 1.

Solution: Multiply.

x2 + ix - ix - i2
x2 + 1 Solving systems of equations graphically is one of the
easiest ways to solve systems of simple equations
(it's usually not very practical for complex equations
such as hyperbolas or circles). One 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. 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). Cramer's Rule is when you use determinats to solve a system of linear equations The last method, Elimination, 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. Method 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 be-
cause 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).
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) 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. Problem: Graph the following system:
2x + y >= 2
4x + 3y <= 12
(1/2) <= x <= 2
y >= 0 When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
This can be explained with a formula: (y2 - y1)/(x2 - x1). To find the slope, you pick any two points on the line and find the change in y, and then divide it by the change in x. Problem: The points (1,2) and (3,6) are on a line.
Find the line's slope.

Solution: Plug the given points into
the slope formula.

y2 - y1
m = -------
x2 - x1

6 - 2
m = -----
3 - 1

After simplification, m = 2 A function is a relation (usually an equation) in which no two ordered pairs have the same x-coordinate when graphed. One way to tell if a graph is a function is the vertical line test, which says if it is possible for a vertical line to meet a graph more than once, the graph is not a function. The figure above is an example of a function Functions are usually denoted by letters such as f or g. If the first coordinate of an ordered pair is represented by x, the second coordinate (the y coordinate) can be represented by f(x). In the figure below, f(1) = -1 and f(3) = 2.
When a function is an equation, the domain is the set of numbers that are replacements for x that give a value for f(x) that is on the graph. Sometimes, certain replacements do not work, such as 0 in the following function: f(x) = 4/x (you cannot divide by 0). In that case, the domain is said to be x <> 0.

There are a couple of special functions whose graphs you should have memorized because they are sometimes hard to graph. They are the absolute value function Graphs of quadratic functions are called parabolas. The basic graph that you need to know is f(x) = x2.
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