In geometry two triangles are similar if the corresponding sides are in proportion. If two angles are the same, the are also similar. If two angles are the similar, in fact, all three angles must be the same.

For example

In triangle ABC, AB = 10, BC = 15, AC = 18

In triangle DEF, DE = 20, EF = 30, DF = 36

These triangles are similar because corresponding sides are in proportion

10/20 = 15/30 = 18/36. All three fractions simplify to 1/2

Try another example.

Triangle ABC, AB = 5, BC = 12, AC = 15

Triangle DEF, DE = 15, BC = 30, AC = 45

For the triangles to be similar, 5/15 = 12/30 = 15/45

Simplify each fraction to get 1/3 = 2/5 = 1/3.. Notice this is a false statement, so the triangles are not similar

Another example dealing with the angles

Triangle ABC, <A = 45, <B = 89

Triangle DEF, <E = 89, <F = 46

<C = 180 - (45 + 89) = 46

<D = 180 - (46 + 89) = 45

<A corresponds with <D, <B corresponds with <E and <C corresponds with <F. Since the angles are congruent, the triangles are similar.

## Sunday, March 31, 2013

## Wednesday, March 27, 2013

ax^2 + bx + c = 0

ax^2 + bx = -c (Subtract c from both sides of the equation)

x^2 + (b/a)x = -c/a (Divide both sides of the equation by a)

x^2 + (b/a)x + b^2/(4a^2) = -c/a + b2/(4a^2) (Complete the square by taking half of (b/a) , squaring it and adding to both sides)

[x + (b/2a)]^2 = -c/a + b^2/(4a^2) (Factor the perfect square trinomial on the left side)

[x + (b/2a)]^2 = (b^2 – 4ac)/4a^2 (Get common denominator of 4a^2 on the right side)

x + (b/2a) = +/- √[(b^2 – 4ac)/4a^2] (Take the square root of both sides (use the square root property))

x + (b/2a) = +/- √(b^2 – 4ac)/2a (Simplify the square root on the right side √(4a^2) = 2a)

x = [-b +/- √(b^2 – 4ac)]/2a (Subtract b/2a from both sides)

The result is known as the quadratic formula

## Saturday, March 23, 2013

For example.. 5 + 2 = 7. To undo the addition of 2 to 5, subtract 2 from 7. 7 - 2 = 5

8 - 5 = 3. To undo the subtraction, add 5 to 3 to get 3 + 5 = 8

Now try multiplication to undo division. 6 divided by 3 = 2. To undo, multiply 2 and 3 to get 6.

This works, but in the next example, it will not.

4 times 0 = 0. If you follow what was done in the examples above, 0 divided by 0 should equal 4, but that is obviously not true.

## Thursday, March 21, 2013

^{2}= -2 is not a real number solution because the square of a real number is always positive. In fact, when trying to find the square root of a negative number on a calculator, you will get a "syntax error", or simply "error" message. Therefore, to solve such equations, the number "i" was created such that i

^{2}= -1. The imaginary number "i" is defined as √-1 = "i". The imaginary numbers are part of another number system known as the complex number system.

To solve the square root of a negative number or expression, we use the same rules for multiplying and dividing radicals. We write the square root of a negative as "i" times a real number. In general, for any real number a > 0, √-a = i√a.

**Examples:**Simplify each of the following radicals.

1. √-16 = √-1√16

= i∙4 (Replace √-1 with i)

= 4i

2. √-5 = √-1√5

= i√5 (Replace √-1 with i)

=√5i

Note that i√5 and √5i are the same. Generally, we write the imaginary number with the radical after the "i". The reason is because it's easy to confuse √(5i) and √5i.

3. √-24 = √-1√24

= √-1√4√6 (Break √24 down into √4√6)

= i∙2√6 (Simplify √4 and Replace √-1 with i)

=2i√6

4. √(-36/81) = √-36 /√81 (Rewrite as a division of two radicals)

= (√-1√36)/√81

= i√36/√81 (Replace √-1 with i)

= 2i/3 (Simplify 6/9 to 2/3)

In the previous examples, we showed how to simplify the square root of negative real numbers. We can extend this to simplifying the square root of monomials containing a negative real number and variables.

**Examples:**

1. √-49x

^{2}y

^{4}= √-1 ∙√49∙√x

^{2}∙√y

^{4}(Rewrite as product of radicals)

= i∙7xy

^{2}(Simplify all radicals and replace √-1 with i)

= 7i(xy

^{2})

2. √-7x

^{3}y

^{2}= √-1∙√7∙√x

^{3}∙√y

^{2}(Rewrite as product of radicals)

= i∙√7∙x∙√x∙y (Simplify all radicals and replace √-1 with i)

= xyi√(7x)

We learned how to simplify the square root of negative numbers with the use of the imaginary number "i". Next we use the imaginary number to define complex numbers. A complex number is any number of the form a + bi where a and b are real numbers and √-1 = "i". Some examples are complex numbers are as follows:

4 - 13i, -12 + 27i, 3 - 3i√2 and (5/2) + (3/2)i.

We can add, subtract, multiply and divide complex numbers. To add and subtract complex numbers, combine the real parts and the imaginary parts.

**Examples:**Add or subtract the following complex numbers.

1. (3 + 4i) + (10 + 6i)

3 + 10 = 13 (Add the real parts)

(4 + 6)i = 10i (Add the imaginary parts)

34

Therefore, (3 + 4i) + (10 + 6i) = 13 + 10i.

2. (-14 + 3i) - (4 + 7i)

-14 - 4 = -18 (Subtract the real parts)

(3 - 7)i = -4i (Subtract the imaginary parts)

Therefore, (-14 + 3i) - (4 + 7i) = -18 - 4.

Recall the product rule when multiplying two positive real numbers. If a and b are positive real numbers, then √a√b = √(ab). When trying to apply the same rule to multiplying the square root of two negative numbers, we get a wrong result. For example, using the product rule for √-3√-27 gives us √(-3)(-27) = √81 = 9. But this is incorrect. What we need to do is write each radical in terms of "i". When doing this we get i√3 ∙ i√27 = (i2)√81 = (-1)(9) = -9.

**Examples:**Multiply the following complex numbers.

1. √-40√-10 = √-1√40√-1√10

= i∙√40∙i∙√10 (Replace √-1 with i)

= i2∙ √400 (Multiply the i's and use product rule for radicals to multiply √40√10)

= (-1)(20) (Replace i

^{2}with -1 and √400 with 20)

= -20

2. √-36√-9 = √-1√36√-1√9

= i∙6∙i∙3 (Simplify the radicals and replace √-1 with i)

= i2∙18 (Multiply the i's and multiply 6 and 3)

= (-1)(18) (Replace i2 with -1)

= -18

The distributive property of multiplication and FOIL that apply to real numbers also apply to complex numbers. Sometimes a problem is in the form (a - bi)(a + bi). These complex numbers are known as complex conjugates and are used in division where the denominator is a complex number. It's used because multiplying complex conjugates results in a real number and that removes the complex number from the denominator. It's the same idea as removing radicals from the denominator. Therefore, to divide complex numbers, multiply the numerator and the denominator by the complex conjugate.

This guide should help anyone having difficulty understanding imaginary numbers and how to perform mathematical operations on complex numbers.

## Saturday, March 16, 2013

To graph f(x) = √x, we'll set up a table of values.

For x = 1, f(1) = √1 = 1

x = 4, f(4) = √4 = 2

x = 9, f(9) = √9 = 3

x = 16, f(16) = √16 = 4

We would plot this on a rectangular coordinate system with the horizontal axis being x, the vertical axis being f(x). Note that the vertical axis on a rectangular coordinate system is generally the y axis. Notice we used values for x that are perfect squares. They are much easier to graph.

Note that we only used the principal square root for the values of f(x). The reason for this will be noted

after the graph. Recall that the square root has both positive and negative values (ie, √4 = 2 and -2). But the reason we did not use the negative values on the graph is because we are assuring that it is a function. If we graph the negative values, the graph is a parabola which opens to the right. This would not pass the vertical line test and would not be a function. Therefore the graph of the function f(x) = √x are only the values in the first quadrant.

Knowing the graph of the basic square root function, f(x) = √x, will help us graph many other radical functions. If h > 0, then f(x) = √(x +h) and f(x) = √(x - h) are the graphs of f(x) = √x moved to the left and right h units, respectively. For example, the graph of f(x) = √(x + 1) has an x-intercept is (-1, 0). When x = 0, f(x) = 1. The domain is x ≥ -1 because x < -1 will yield a negative under the radical and the square root of a negative is not a real number. You can plot several other points that makes √(x + 1) a perfect square. This graph is the graph of f(x) = √x moved 1 unit to the left. It is said to be translated 1 unit to the left. The graph of f(x) = √(x - 1), similarly, would be the graph of f(x) = √x moved 1 unit to the right.

We learned how to graph radical functions that are translations left and right of the square root function, f(x) = √x. Now we will learn how to graph radical functions that are that translations up and down of the square root function f(x) = √x. If k >0 , then f(x) = √x - k and f(x) = √x + k are the graphs of f(x) moved down k units and up k units, respectively. Therefore the graph of f(x) = √x + 3 is the same as the graph of f(x) = √x except it is translated 3 units up. Similarly, the graph of f(x) = √x - 3 is the same as the graph of f(x) = √x except it is translated 3 units down.

A function can have multiple translations. For example, the graph of f(x) = √(x - 2) + 1 is the same as the graph of f(x) = √x except it is translated 2 units right and 1 unit up. You will notice when graphing that there is vertical line through x = 2. The domain of the function is x ≥ 2. This shows that there are no points on the graph to the left of 2 on the x axis. The graph is in the first quadrant where both x and f(x) are positive.

The cube root function is a radical function defined as f(x) =

^{3}√x. To graph the function we want to choose values for x that are a perfect cube. For example 1 is a perfect cube because 1(1)(1) = 1 and 8 is a perfect cube because (2)(2)(2) = 8.

To graph f(x) =

^{3}√x, we'll set up a table of values.

Fox x = 1, f(1) =

^{3}√(1) = 1

x = -1, f(-1) =

^{3}√(-1) = -1

x = 8, f(8) =

^{3}√(8) = 2

x = -8, f(-8) =

^{3}√(-8) = -2

x = 27, f(27) =

^{3}√(27) = 3

x = -27, f(-27) =

^{3}√(-27) = -3

The function contains values in the first and third quadrants and no values in the second and fourth

quadrants. As with the square root function, there are translations with the cube root function. The graph of f(x) =

^{3}√x - 3 is the same as the graph of this function is the same the graph of f(x) =

^{3}√x except it is translated down 3 units.

This guide should help clear any confusion on the topic of radical functions and graphing radical functions.

## Tuesday, March 12, 2013

Absolute value is noted as follows:

Absolute value of 4... |4| = 4.

Absolute value of -4... |-4| = 4.

When solving equations with absolute value, we have to set up two equations, a positive case and a negative case. For example, |x| = 5 means that the positive case is x = 5 and the negative case is x = -5. The same idea holds true for more complex problems with absolute value.

**Example:**Solve |x + 5| = 9.

The first equation (positive case) is x + 5 = 9, therefore x = 4. The second equation (negative case) is x + 5 = -9, therefore x = -14.

**Example:**Solve |2x + 7| = 4x + 2.

Positive case is 2x + 7 = 4x + 2.

2x = 4x + 2 - 7 (subtract 7 from each side)

2x = 4x - 5

2x - 4x = -5 (subtract 4x from each side to isolate x)

- 2x = -5

x = 5/2 (divide both sides by -2)

Negative case is 2x + 7 = -(4x + 2)

2x + 7 = -4x - 2 (apply the - in front of the parentheses on the right side)

2x = -4x - 2 - 7 (subtract 7 from both sides)

2x = -4x - 9

2x + 4x = -9 (add 4x to both sides)

6x = -9

x = -9/6 (divide both sides by 6)

x = -3/2 (simplify -9/6 to -3/2)

Substitute each value for x into the original equation to see which satisfy the equation. There is no need for this substitution on equations without a variable on both sides of the equation. In this case, x = -3/2 will not satisfy the equation and therefore is not a solution.

**Example:**Solve 3|4x - 5| = 6x.

Positive case is 3(4x - 5) = 6x

12x - 15 = 6x (apply the distributive property)

12x = 6x + 15 (add 15 to both sides)

12x - 6x = 15 (subtract 6x from both sides)

6x = 15 (combine like terms on left side)

x = 15/6 (divide both sides by 6)

x = 5/2 (simplify 15/6 to 5/2)

Negative case is 3(4x - 5) = -6x

12x - 15 = -6x

12x = -6x + 15

12x + 6x = 15

18x = 15

x = 15/18

x = 5/6

When check the possible solutions to make sure they satisfy the original equation, you will see that both satisfy the equation, therefore they are both solutions.

The absolute value inequality will be in the forms |x| < h, |x| ≤ h, |x| > h or |x| ≥ h. To solve such equations, we still have a positive and a negative case. To set up the equations for the positive case, we drop the absolute value sign and rewrite the equation . For the negative case, we drop the absolute value sign, take the negative of the other side and reverse the inequality symbol.

**Example:**Solve |2x - 5| < 9.

Positive case 2x - 5 < 9

2x < 14

x < 7

Negative case 2x - 5 > -9

2x > -4

x > -2

The solution is the set (-2, 7).

**Example:**Solve |4x - 3| ≥ 13.

Positive case 4x - 3 ≥ 13

4x ≥ 16

x ≥ 4

Negative case 4x - 3 ≤ -13

4x ≤ -10

x ≤ -10/4

x ≤ -5/2

The solution is (-∞, -5/2] U [4, ∞).

The most important thing to remember is to set up both positive and negative cases. In the negative case with inequalities, remember to flip the inequality sign and negate whatever is after the sign.

This guide should help clear any confusion about absolute value and solving equations and inequalities dealing with absolute value.

## Saturday, March 9, 2013

When solving for the volume of a sphere, cylinder, cone or solving for the circumference or area of a circle, the answer is typically left in terms of Pi.

If you wish to have a numerical value, you can use the approximation 22/7 for Pi or use 3.14 for Pi. Of course, if you want more accuracy with your result, you can use more decimal places for Pi. Use as many as you wish.

## Thursday, March 7, 2013

Tan(x + pi) - cos (x + pi/2) =0

[Tan(x)+Tan(pi)]/(1 - tan(x)tan(pi)] - [ cos(x)cos(pi) - sin(x)sin(pi/2) ] = 0

[tan(x) + 0]/(1 - tan(x)(0)] - [ cos(x)(0) - sin(x)(1)] = 0

tan(x) + sin(x) = 0

tan(x) = -sin(x)

sin(x)/cos(x) = -sin(x)

sin(x) = -sin(x)cos(x)

sin(x) + sin(x)cos(x) = 0

sin(x)[1 + cos(x)] = 0

sin(x) = 0, 1 + cos(x) = 0

x =0 cos(x) = -1

x = pi

sin(x)/cos(x) = -sin(x)

sin(x) = -sin(x)cos(x)

sin(x) + sin(x)cos(x) = 0

sin(x)[1 + cos(x)] = 0

sin(x) = 0, 1 + cos(x) = 0

x =0 cos(x) = -1

x = pi

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