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Introduction To Algebraunknowns : equations : simultaneous : quadratic 
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In order to learn Algebra, you need to find unknown numbers from information given. Here are two examples of the type of problems that can be solved using algebra.
Example 1: When six is added to this number, it gives nine. What is the number?
Example 2: A number is multiplied by itself. Three times that same number is subtracted. Finally, two is added. The total is now zero. What is the number?
Example 1 is trivial. The number has to be three. Example 2 is much harder to guess. I will return to it later.
Two numbers added together are shown as:
Two numbers subtracted are shown as:
Two numbers multiplied together are shown as:
but more commonly:
Two numbers divided are shown as:
A number multiplying a sum of numbers is shown as:
This can be expanded by multiplying everything inside the brackets by the number outside:
A number dividing a sum of numbers is shown as:
This can be expanded by dividing everything inside the brackets by the number outside:
Two positive numbers multiplied together give a positive number:
Two negative numbers multiplied together also give a positive number:
A negative number multiplied by a positive number gives a negative number:
This is called an equation because there is an equals sign. If we want to find the value of the unknown number, X, we have to get the X on one side of the equation and the numbers on the other side.
We can do whatever we like to this equation as long as we do the same to both sides of the equation.
We can easily isolate the X by subtracting 6 from both sides:
So the unknown number is 3. It is always a good idea to check an answer by puting the value 3 back in the original equation:
Checking by replacing X with 3 gives: 3 + 6 = 9 ..... correct.
Example 2 from above will be dealt with later.
Example 3: Solve the equation X  6 = 2.
Again, we have to isolate the X on one side of the equation and the numbers on the other side. This can be done by adding 6 to both sides.
Checking by replacing X with 8 gives: 8  6 = 2 ..... correct.
Example 4: Solve the equation 2X  3 = 1
This one is a little more complicated. The X is multiplied by 2 and then 3 is subtracted. We must deal with the 3 being subtracted first. We can do this by adding 3 to both sides.
The next step is to divide both sides by 2.
Checking by replacing X with 1 gives: (2 × 1)  3 = 2  3 = 1 ..... correct.
Example 5: Solve the equation, 2(X + 3) = 4.
In this equation, the 2 multiplies everything inside the bracket. This 2 must be dealt with first. It is removed by dividing both sides by 2.
Now we can subtract 3 from both sides.
Checking by replacing X with 1 gives: 2 × (1 + 3) = 2 × 2 = 4 ..... correct.
Example 6: Solve the equation 4 / (X + 1) = 2.
In this equation the part containing the X is at the bottom. The X + 1 in the bracket is all dividing the 4. It must be brought to the top. This is done by multiplying both sides by (X + 1).
The next step is to divide both sides by 2.
Subtract 1 from both sides.
Checking by replacing X with 1 gives: 4 / (1 + 1) = 4 / 2 = 2 ..... correct.
Example 7: Solve the equation X  3 = 2.
The X has a minus sign in front of it. It is being multipied by 1:
To remove the 1 we need to multiply both sides by 1.
Multiplying by 1 is the same as changing the signs throughout. The next step is to subtract 3 from both sides.
Checking by replacing X with 1 gives: (1)  3 = 1  3 = 2 ..... correct.
Note that two minuses multiplied together make a plus.
Example 8: Solve the following equation, 2X  5 = 4  X.
The usual rules apply: get the X on one side and the numbers on the other side. Add X to both sides to remove the X from the right hand side.
Two Xs and a single X can be added to give 3 Xs.
Add 5 to both sides.
Divide both sides by 3.
Checking by replacing X (in 2X  5) with 3 gives: (2 × 3)  5 = 6  5 = 1
Checking by replacing X (in 4  X) with 3 gives: 4  3 = 1 ..... both sides agree.
X = 1 and Y = 2 (because 1 + 2 = 3).
But that is not the only answer.
X = 0 and Y = 3 (because 0 + 3 = 3).
X = 1 and Y = 4 (because 1 + 4 = 3).
X = 4 and Y = 1 (because 4 + 1 = 4  1 = 3).
In fact, an equation with two unknowns has an infinite number of pairs of answers. To fix a single pair of numbers as the answer, we need another equation.
A pair of equations, each with two unknown numbers are called Simultaneous Equations. They can be solved together to give the values for the unknowns that satisfy BOTH equations simultaneously (hence the name).
Example 9: Solve the simultaneous equations: X + Y = 3 and X  Y = 7.
Firstly we list and number the two equations:
X + Y = 3  ..... (i) 
X  Y = 7  ..... (ii) 
Next we manipulate the two equations to produce a new equation that eliminates either X or Y from one of the equations. In the above example we can do that by adding the two equations (i) and (ii).
This gives us a value for X. By replacing X in (i) with 5 we can get a value for Y.
So we have a pair of values, X = 5 and Y = 2. These can be checked by putting them into (ii).
Checking the values, X = 5, Y = 2 in (ii) gives: X  Y = 5  (2) = 5 + 2 = 7 ..... correct.
Example 10: Solve the simultaneous equations: 2X + 3Y = 4 and X  Y = 3.
List and number the two equations:
2X + 3Y = 4  ..... (i) 
X  Y = 3  ..... (ii) 
A simple addition or subtraction is not sufficient to eliminate either X ot Y. We need to begin by doubling (ii).
2X + 3Y = 4  ..... (i) 
2X  2Y = 6  ..... (ii) × 2 .... (iii) 
We can now subtract the two equations.
We now have our value for Y. By replacing Y in (i) with 2 we can get a value for X.
Again, we have a pair of values, X = 1 and Y = 2. These can be checked by putting them into (ii).
Checking the values, X = 1, Y = 2 in (ii) gives: X  Y = 1  (2) = 1 + 2 = 3 ..... correct.
Example 2: A number is multiplied by itself. Three times that same number is subtracted. Finally, two is added. The total is now zero. What is the number?
This can be written in equation form as:
An equation where X squared occurs with X and numbers is called a Quadratic Equation. Quadratic Equations generally have two answers rather than one. In other words, there are two values of X that can satisfy the equation.
There are two methods of solving Quadratic Equations. One is called factorisation. An equation of the form
can be broken into a form like
This process is trial and error. For the equation, X^{2}  3X + 2 = 0 this becomes
Note: When you multiply out the two brackets, you multiply each term in the first bracket by each term in the second bracket:
More algebraic expansions can be found in the Pascals' Triangle essay.
Let us look again at (X  2)(X  1) = 0. This is a multiplication of the form
When two numbers are multiplied together and give zero, either G must be zero or H must be zero.
So, when (X  2)(X  1) = 0, either X  2 = 0 or X  1 = 0.
The first is satisfied when X = 2, the second when X = 1. These are the two values for the equation X^{2}  3X + 2.
Checking by replacing X with 2 gives: 2^{2}  (3 × 2) + 2 = 4  6 + 2 = 0 .... correct.
Checking by replacing X with 1 gives: 1^{2}  (3 × 1) + 2 = 1  3 + 2 = 0 .... correct.
A better method of solving Quadratic Equations is by using the formula below, discovered by Omar Khayyam in the 12th century.
When ax^{2} + bx + c = 0,
This is the Quadratic Formula. There are two solutions because of the presence of the plus and minus options.
For the equation, X^{2}  3X + 2,
A = 1, B = 3, C = 2.
Using the Quadratic Formulas above we get:
Again we end up with X = 2 and X = 1.
Example 11: Solve the Quadratic Equation, X^{2}  5X + 6 = 0.
The coefficients are A = 1, B = 5, C = 6.
Using the Quadratic Formulas we get:
We end up with X = 2 and X = 3.
Checking by replacing X with 2 gives:
2^{2}  (5 × 2) + 6 = 4  10 + 6 = 0 .... correct.
Checking by replacing X with 3 gives:
3^{2}  (5 × 3) + 6 = 9  15 + 6 = 0 .... correct.
Example 12: Solve the Quadratic Equation, (X  1)(X + 2) = 1  X  2X^{2} .
This equation needs to be rearranged. Firstly we multiply out the left hand side:
(X  1)(X + 2) which expands to X^{2} + X  2
The full equation now becomes
Getting every term onto the left hand side:
A = 3, B = 2, C = 1.
Using the Quadratic Formulas we get:
We end up with X = 1 and X = 1 / 3.
Checking by replacing X with 1 gives:
3 × 1^{2} + (2 × 1)  1 = 3  2  1 = 0 .... correct.
Checking by replacing X with 1 / 3 gives:
3 × (1 / 3)^{2} + (2 × (1 / 3))  1 = 1 / 3 + 2 / 3  1 = 0 .... correct.
Remember to always check your results.
They are a special case of the Quartic Equation. These are essentially Quadratic Equations in x^{2}.
Example 13: Solve the equation X^{4}  3X^{2} + 2 + 0.
We can use the Quadratic formula by putting A = 1, B = 3, C = 2:
This gives X^{2} = 2 and X^{2} = 1. This means that there are four solutions, which is correct for a Quartic Equation:
The part inside the square root sign is called the Determinant. Its value determines the type of roots.
Complex Numbers are of the form a + bi where a and b are real numbers and i = √ 1.
Example 14: Solve the equation X^{2} + X + 1 + 0.
Using the Quadratic formula and putting A = 1, B = 1, C = 1:
This gives X = 1/2 + i √ 3/2 and X = 1/2  i √ 3/2.
A pair of complex roots of the form a + ib and a  ib are called Complex Conjugates. If a Quadratic Equation has complex roots they will always be a pair of Compex Conjugates.
In this section we will look at the roots of two types of Cubic Equations.
The first type are called Pure Cubic and take the form X^{3}  T = 0, where T is any real number.
The roots of this type are given by
x = ω_{2} ^{3}√ T
x = ω_{3} ^{3}√ T
ω_{3} = (1  i √ 3) / 2
The second type is the Symetric Cubic: X^{3} + PX^{2} + PX + 1 = 0.
This can be factorised to (X + 1)(X^{2}  X + 1 + PX) = 0.
This gives a value of x = 1. The remaining Quadratic can be solved using the Quadratic Formula.
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