algebra 2

Inequalities a < b      means a is less than b (so b is greater than a) a £ b      means a is less than or equal to b (so b is greater than or equal to a) a ³ b      means a is greater than or equal to b etc. a > b      means a is greater than b etc. If you have an inequality, you can add or subtract numbers from each side of the inequality, as with an equation. You can also multiply or divide by a constant. However, if you multiply or divide by a negative number, the inequality sign is reversed. Example Solve 3(x + 4) < 5x + 9 3x + 12 < 5x + 9\ -2x < -3\x > 3/2   (note: sign reversed because we divided by -2) Inequalities can be used to describe what range of values a variable can be. E.g. 4 £ x < 10, means x is greater than or equal to 4 but less than 10.Graphs Inequalities are represented on graphs using shading. For example, if y > 4x, the graph of y = 4x would be drawn. Then either all of the points greater than 4x would be shaded or all of the points less than or equal to 4x would be shaded. Example: x + y < 7 and 1 < x < 4 (NB: this is the same as the two inequalities 1 < x and x < 4) Represent these inequalities on a graph by leaving unshaded the required regions (i.e. do not shade the points which satisfy the inequalities, but shade everywhere else). Number Lines Inequalities can also be represented on number lines. Draw a number line and above the line draw a line for each inequality, over the numbers for which it is true. At the end of these lines, draw a circle. The circle should be filled in if the inequality can equal that number and left unfilled if it cannot. Example: On the number line below show the solution to these inequalities. -7 £ 2x - 3 < 3This can be split into the two inequalities: -7 £ 2x - 3 and 2x - 3 < 3\ -4 £ 2x and 2x < 6\ -2 £ x and x < 3 The circle is filled in at –2 because the first inequality specifies that x can equal –2, whereas x is less than (and not equal to) 3 and so the circle is not filled in at 3. The solution to the inequalities occurs where the two lines overlap, i.e. for -2 £ x < 3

Matrices are tables of numbers. The numbers are put inside big brackets. Matrices are given 'orders', which basically describe the size of the matrices. The order is the number of rows 'by' the number of columns. So a 2 by 3 matrix has 2 rows and 3 columns. Adding and Subtracting Adding and subtracting matrices is fairly straight-forward. Adding and subtracting of matrices, however, can only occur if the matrices have the same order. MultiplicationThis is a little harder than addition and subtraction. Multiplication cannot occur between any two matrices just like addition can't occur between any two matrices- they have to be compatible. The rule is: if you want to multiply two matrices, if the first matrix has order p×q (p by q) and the second has order y×z, q must equal y or you can't multiply the two matrices. If you can multiply, the answer will be a matrix which has order p×z. Here is how you would multiply a 2×3 and a 3×2 matrix: Example:

A quadratic equation is an equation where the highest power of x is x². There are various methods of solving quadratic equations, as shown below. NOTE: If x² = 36, then x = +6 or -6 (since squaring either of these numbers will give 36). However, Ö36 = + 6 only. Completing the Square 9 and 25 can be written as 3² and 5² whereas 7 and 11 cannot be written as the square of another exact number. 9 and 25 are called perfect squares. Another example is (9/4) = (3/2)². In a similar way, x² + 2x + 1 = (x + 1)². To make x² + 6x into a perfect square, we add (6²/4) = 9. The resulting expression, x² + 6x + 9 = (x + 3)² and so is a perfect square. This is known as completing the square. To complete the square in this way, we take the number before the x, square it, and divide it by 4. This technique can be used to solve quadratic equations, as demonstrated in the following example. Example: Solve x² - 6x + 2 = 0 by completing the square x² - 6x = -2 [To complete the square on the LHS (left hand side), we must add 6²/4 = 9. We must, of course, do this to the RHS also].\ x² - 6x + 9 = 7\ (x - 3)² = 7 [Now take the square root of each side]\ x - 3 = ±2.646     (the square root of 7 is +2.646 or -2.646)\ x = 5.646 or 0.354 Completing the square can also be used to find the maximum or minimum point on a graph. Example: Find the minimum of the graph y = 3x² - 6x - 3 . In this case, the x² has a '3' in front of it, so we start by taking the three out: y = 3(x² - 2x -1) .  [This is the same since multiplying it out gives 3x² - 6x - 3] Now complete the square for the bit in the bracket:\ y = 3[(x - 1)² - 2] Multiply out the big bracket:\ y = 3(x - 1)² - 6 We are trying to find the minimum value that this graph can be. (x - 1)² must be zero or positive, since squaring a number always gives a positive answer. So the minimum value will occur when (x - 1)² = 0, which is when x = 1. When x = 1, y = -6 . So the minimum point is at (1, -6). Some people don't like the method of completing the square to solve equations and an alternative is to use the quadratic formula. This is actually derived by completing the square. The Quadratic Formula Where the equation is ax² + bx + c = 0 Example: Solve 3x² + 5x - 8 = 0 x = -5 ± Ö( 5² - 4×3×(-8))                    6   = -5 ± Ö(25 + 96)                6   = -5 ± Ö(121)              6   = -5 + 11       or        -5 - 11           6                             6 \ x = 1 or -2.33 Factorising Sometimes, quadratic equations can be solved by factorising. In this case, factorising is probably the easiest way to solve the equation. Example: Solve x² + 2x - 8 = 0\ (x - 2)(x + 4) = 0\ either x - 2 = 0 or x + 4 = 0\ x = 2 or x = - 4 If you do not understand the third line, remember that for (x - 2)(x + 4) to equal zero, then one of the two brackets must be zero.

Simultaneous Equations Simultaneous equations are a set of equations which have more than one value which has to be found. At GCSE, it is unlikely that you will have more than two equations with 2 values (x and y) which need to be found. Example: A man buys 3 fish and 2 chips for £2.80 A woman buys 1 fish and 4 chips for £2.60 How much are the fish and how much are the chips? There are two methods of solving simultaneous equations. Use the method which you prefer. Method 1: elimination First form 2 equations. Let fish be f and chips be c. We know that: 3f + 2c = 280    (1) f + 4c = 260      (2) Doubling (1) gives: 6f + 4c = 560 (3) (3)-(2) is 5f = 300\ f = 60 Therefore the price of fish is 60p Substitute this value into (1): 3(60) + 2c = 280\ 2c = 100 c = 50 Therefore the price of chips is 50p Method 2: Substitution Rearrange one of the original equations to isolate a variable. Rearranging (2): f = 260 - 4c Substitute this into the other equation: 3(260 - 4c) + 2c = 280\ 780 - 12c + 2c = 280\ 10c = 500\ c = 50 Substitute this into one of the original equations to get f = 60 . Harder simultaneous equations: To solve a pair of equations, one of which contains x², y² or xy, we need to use the method of substitution. Example: 2xy + y = 10  (1) x + y = 4        (2) Take the simpler equation and get y = .... or x = .... from (2), y = 4 - x    (3) this can be substituted in the first equation. Since y = 4 - x, where there is a y in the first equation, it can be replaced by 4 - x . sub (3) in (1), 2x(4 - x) + (4 - x) = 10\ 8x - 2x² + 4 - x - 10 = 0\ 2x² - 7x + 6 = 0\ (2x - 3)(x - 2) = 0\ either 2x - 3 = 0 or x - 2 = 0 therefore x = 1.5 or 2 . Substitute these x values into one of the original equations. When x = 1.5,  y = 2.5 when x = 2, y = 2 Simultaneous equation can also be solved by graphical methods.

Trial and Improvement Any equation can be solved by trial and improvement (/error). However, this is a tedious procedure. Example: Solve t³ + t = 17 by trial and improvement. Firstly, select a value of t to try in the equation. I have selected t = 2. Put this value into the equation. We are trying to get the answer of 17. If t = 2, t³ + t = 2³ + 2 = 10 . This is lower than 17, so we try a higher value for t. If t = 2.5, t³ + t = 18.125 (too high) If t = 2.4, t³ + t = 16.224 (too low) If t = 2.45, t³ + t = 17.156 (too high) If t = 2.44, t³ + t = 16.966 (too low) If t = 2.445, t³ + t = 17.061 (too high) So we know that t is between 2.44 and 2.445. So to 2 decimal places, t = 2.44. Iteration This is a way of solving equations. An iteration formula might look like the following:xn+1 = 2  + 1                xn You are usually given a starting value, which is called x0. If x0 = 3, substitute 3 into the original equation where it says xn. This will give you x1. (This is because if n = 0, x1 = 2 + 1/x0 and x0 = 3).x1 = 2 + 1/3 = 2.333 333 (by substituting in 3). To find x2, substitute the value you found for x1. x2 = 2 + 1/(2.333 333) = 2.428 571 Repeat this until you get an answer to a suitable degree of accuracy. This may be about the 5th value for an answer correct to 3s.f. In this example, x5 = 2.414... Example: a) Show that x = 1 +  11                                   x - 3 is a rearrangement of the equation x² - 4x - 8 = 0. b) Use the iterative formula Xn+1 = 1 + 11                                                           xn - 3 together with a starting value of x1 = -2 to obtain a root of the equation x² - 4x - 8 = 0 accurate to one decimal place. a) multiply everything by (x - 3): x(x - 3) = 1(x - 3) + 11 so x² - 3x = x + 8 so x² - 4x - 8 = 0 b) x1 = -2x2 = 1 +   11 (substitute -2 into the iteration formula)             -2 - 3     = -1.2x3 = 1 +      11        (substitute -1.2 into the above formula)             -1.2 - 3     = -1.619x4 = -1.381x5 = -1.511x6 = -1.439x7 = -1.478 therefore, to one decimal place, x = 1.5 .

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