Department of Economics Maths Revision Notes 4

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1 Department of Economics Maths Revision Notes 4 1. Quadratic expressions Expanding (multiplying out) pairs of brackets Consider an expression such as (a + b)(c + d) This is a product - that is, (c + d) multiplied by (a + b). We can think of this product as being the area of a rectangle with one side of length (a + b) and the other side (c + d), as in Fig. 1. a + b This area = (a + b)(c + d) c + d Fig. 1 But, we see from Fig that the total area is composed of four sub-areas: ac, ad, bc, and bd. a b ac bc c Fig. Therefore the total area is ad bd d = (a + b)(c + d) (from Fig. 1) = ac + ad + bc + bd (from Fig. ) = a(c + d) + b(c + d) In words, the second bracket is multiplied by each of the terms in the first bracket, and the results added.

2 So the rule is: Rule 1: (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd Caution: when applying this rule, be careful to obey the sign rules. For example, (a - b)(c + d) = a(c + d) - b(c + d) = ac + ad - bc - bd A special case of Rule 1 arises when a = c and b = d. Then the rule becomes: Rule 1a: (a + b)(a + b) = (a + b) = a(a + b) + b(a + b) = a + ab + ba + b = a + ab + b In our discussion up to this point, we have left open the question whether a and b are variables or constants, as this has no effect on the reasoning. However, we shall frequently meet expressions of the form (x + a)(x + b) where x is a variable or unknown, and a and b are constants. Applying Rule 1 to this expression, we get Rule 1c: (x + a)(x + b) = x(x + b) + a(x + b) = x + xb + ax + ab = x + (a + b)x + ab It is worth taking careful note of the pattern in this last expression, as we shall meet it frequently. In the second term, the coefficient of x is the sum of the two constants, a and b. The third term, which is a constant since it does not contain x, is the product of the two constants, a and b.

3 Note also that Rule 1c implies that (x + a)(x + b) and x + (a + b)x + ab are identically equal (i.e. irrespective of the values of x, a, and b). Finally, note that x + (a + b)x + ab is the standard form of what is called a quadratic expression. A quadratic expression contains an unknown (x) raised to the power (and no higher). It is also known as an equation of the second degree. Factorising quadratic expressions We have just seen that if we expand (multiply out) the expression we get (x + a)(x + b) x + (a + b)x + ab Factorising a quadratic expression is simply the reverse of this process; that is, we start with x + (a + b)x + ab and deduce that the factors of this expression are (x + a) and (x + b). Technique for factorising a quadratic expression Example 1: Suppose we are asked to factorise x + 9x + 0 If we compare this with the standard form, from above: x + (a + b)x + ab we see that a + b = 9, and ab = 0. So we are looking for two numbers, a and b, whose sum is 9 and whose product is 0. After a little thought, it occurs to us that a = 5 and b = 4 satisfy these requirements, for then we have a + b = = 9 and ab = 5 x 4 = 0 Therefore the factors are x + 5 and x + 4. We can check that this is correct by expanding (x + 5)(x + 4) which by Rule 1 gives

4 x + 4x + 5x + 0 = x + 9x + 0 which is what we were asked to factorise. Example : x - x - 4 Again we compare this with the standard form, from above: x + (a + b)x + ab and we see that this time we have a + b = -, and ab = -4. This case is a little trickier because of the minus signs. But we can deduce that a and b must have opposite signs (that is, one positive, the other negative), because their product, ab, is negative. So let us assume that a is negative and b positive. Also, if a is negative and b positive, then a must be greater in absolute magnitude than b, because their sum, a + b, is negative. Using these inferences, we can quickly list the possible values of a and b, given that their product must be -4: a b ab a + b Clearly it is only the last of these that gives the right values for both ab and a + b. So we conclude that the factors are x - 6 and x + 4. We check this by expanding (x - 6)(x + 4) = x + 4x -6x - 4 = x -x - 4 which is what we were asked to factorise. Caution: Not every quadratic expression can be factorised. Later we shall see exactly what conditions must be fulfilled in order that a quadratic expression may be factorised. Clearly, these methods of factorising quadratic expressions are rather unsatisfactory, because they rely on essentially "trial and error" methods. Fortunately, there is a much more powerful method of factorising quadratic expressions, which we examine shortly. Meanwhile, you may be wondering why we should want to factorise quadratic expressions. This will become clear in the next section.

5 3 Quadratic equations 3.0 Suppose we were asked to solve the equation x + 9x + 0 = 0 (1) This is an example of a quadratic equation. (Previously, we considered quadratic expressions - these contained no = sign). To solve it, we factorise the left hand side. By trial and error, we find that the factors are x + 5 and x + 4. So we know that x + 9x + 0 = (x + 5)(x + 4) Moreover, as noted earlier, this equation is an identity - that is, it is true for all values of x. This fact permits us to replace x + 9x + 0 in equation (1) with something identical, namely the product (x + 5)(x + 4). This gives (x + 5)(x + 4) = 0 () The key point is that since equation () is identical to equation (1), it follows that any value or values of x that satisfy () will automatically satisfy (1). So now we want to solve equation (). Why is () any easier to solve than (1)? The answer is that the left hand side of () is a product, so we have something of the form A x B = 0 where A = (x + 5) and B = (x + 4) Now, the product A x B equals zero whenever either A = 0 or B =0 (since "zero times anything is zero"). Similarly, (x + 5)(x + 4) equals zero whenever either (x + 5)= 0 or (x + 4) = 0 So, equation (): (x + 5)(x + 4) = 0 () is satisfied when either (x + 5)= 0 (which is true when x = -5) or (x + 4) = 0 (which is true when x = -4) Therefore we conclude that x = -5 and x = -4 are solutions to equation (1).

6 We can check that these solutions are correct by substituting each of them in turn into equation (1). (You should do this as an exercise). So the solutions (or roots, as they are usually called) to the quadratic equation x + 9x + 0 = 0 are x = -5 and x = -4 Generalisation To solve any quadratic equation of the form ax + bx + c = 0 (where a, b, and c are given constants) we first factorise the quadratic expression on the left hand side of the equation. Let us assume that we find the factors to be Then we have x + α and x + β (x + α)(x + β) = ax + bx + c = 0 Then the roots (solutions) of the given quadratic equation are x = -α and x = -β. (Notice the minus signs in front of α and β. Recall, we need x = -α to make (x + α) = 0 and x = -β to make (x + β) = 0.) 3.1 The formula for solving quadratic equations In 3.1 above we factorised quadratic expressions by "trial and error" methods. This was not very satisfactory. Fortunately, a formula exists (which we will not derive here) for solving quadratic equations. This takes us straight to the roots, however diffficult the numerical values of the parameter, and also tells us whether a solution exists. The formula is: Rule 9: Given any quadratic equation ax + bx + c = 0 (where a, b, and c are given constants) the solutions (roots) are given by x = b ± b 4ac a

7 Notice in the formula the symbol "±", meaning "plus or minus". Thus the formula generates the necessary two roots, as we take first the "plus" option, then the "minus". This will become clearer in the example below. This formula looks rather daunting at first sight, but with a little practice you should have no difficulty in remembering and applying it correctly. Example: x + 5x + 6 = 0 Comparing with the standard form (see Rule 9), we see that a = 1, b = 5, and c = 6. Substituting these values into the formula, we get x = b ± b 4ac a = 5 ± = 5 ± 5 4(1)(6) (1) 1 = - or -3 = either 5+1 or 5 1 = 5 ± 5 4 (note that 1 = ±1 ) These solutions (roots) can be checked by substituting each in turn into the given equation. Notice, incidentally, that these solutions imply that the factors of x + 5x + 6 are x + and x + 3. We can confirm this by expanding (x + )(x + 3) which gives (x + )(x + 3) x + 3x + x + 6 x + 5x Cases where a quadratic expression cannot be factorised In 4..1 above we said that not all quadratic expressions could be factorised. Now we can see why. Consider for example x + x + = 0 Let us see what happens when we try to solve the same equation using the formula. In this case we have a = 1, b = 3, and c = 9. Substituting these values into the formula gives x = b ± b 4ac a = ± 4(1)() (1) = ± 4 8 = ± 4

8 At this point we grind to a halt. To go any further requires us to find the square root of minus 4, and we know from Revision Notes 1 that a negative number has no square root. Thus we conclude that the equation x + x + = 0 has no solutions (roots). 1 Generalisation. The case above arose because, in the formula, the expression under the square root sign was negative. When we look again at the formula x = b ± b 4ac a it is clear that this problem will arise whenever the parameters of the equation ax + bx + c = 0 (where a, b, and c are given constants) are such as to make b less than 4ac. (You should find it easy to invent examples where this occurs) 3.3 The case of the "perfect square" Consider, for example, the quadratic equation x + 6x + 9 = 0 Let's see what happens when we try to solve this, using the formula. In this case we have a = 1, b = 6, and c = 9. Substituting these values into the formula gives x = b ± b 4ac a = 6 ± 6 4(1)(9) (1) = 6 ± = 6 = -3 Thus the equation has only one solution (root), -3. This happens because the expression under the square root sign happens, in this case, to equal zero (and the square root of zero is zero). It therefore disappears ("drops out") of the solution, leaving only one solution. You might think this means that x + 6x + 9 has only one factor, (x + 3). But this can't be right, since we must multiply (x + 3) by something in order to produce 1 Actually this conclusion is not quite correct, as it stands. Mathematicians, who have very tidy minds, found it very unsatisfactory that some quadratic equations had solutions, and some did not. They therefore defined a new type of number, called a complex number, which has the property that its square is negative. (See also Revision Notes ) We won't go into the details here, but this definition makes it possible to conclude that the equation above does, in fact, have roots, but these roots are not real numbers, but rather are complex numbers. Thus the correct conclusion is the the equation above has no real roots.

9 x + 6x + 9. That something is (x + 3). In other words, we have (x + 3)(x + 3) = x + 6x + 9 The expression x + 6x + 9 is called a perfect square, because it corresponds to the area of a square with sides of length (x + 3). Generalisation. The case above arose because, in the formula, the expression under the square root sign was zero. When we look again at the formula x = b ± b 4ac a it is clear that this problem will arise whenever the parameters of the equation ax + bx + c = 0 (where a, b, and c are given constants) are such as to make b equal to 4ac. 3.4 Summary Given the quadratic equation ax + bx + c = 0 (where a, b, and c are given constants) the solutions (roots) are given by x = b ± b 4ac a There are three classes of solution, depending on the values of the parameters, a, b, and c: (i) (ii) (iii) If b > 4ac, there are two (real) roots. If b = 4ac, there is only one (real) root, described as a repeated root. If b < 4ac, there are no roots (more precisely, there are no roots that are real numbers, but roots may be found by defining a new type of number, a complex number.

10 4. Quadratic Functions If we write, for example y = x + 5x + 6 we have a function (that is, a relationship between an independent variable, x, and a dependent variable, y. Because x is raised to the power (and to no higher powers), this is called a quadratic function. The general form of the quadratic function is y = ax + bx + c (where a, b, and c are parameters) Let's consider what the graph of a quadratic function will look like. Example 1 We will start with the simplest case, y = x which of course is obtained by setting a = 1 and b = c = 0 in the general form. If we draw up a table of values for values of x between -4 and +4, and plot the resulting values as a graph, the result is seen in Fig. 1 x y = x (-4) =16 (-3) =9 (-) =4 (-1) = Notice that y is always positive, irrespective of whether x is positive or negative. Fig. 1: y 16 y = x x

11 This curve is called a parabola. The U-shape arises from the fact that x (and therefore y) is positive when x is either positive or negative (for example, y = 4 when x = + and also when x = -). Therefore the curve is symmetrical about the y-axis. This curve gives us the opportunity to introduce the concept of single-valued and multivalued functions. To any value of x there is one and only one associated value value of y. (For example, when x = 3, y = 9). We therefore say that y is a single-valued function of x. However the converse is not true. To any positive value of y there are two corresponding values of x (for example, when y = 9, x = +3 or -3). We therefore say that x is a double-valued function of y. Example y = -x This is the same as Example 1 except that we now have a = -1 instead of +1 in the general form. Compared with Example 1, the sign of y associated with any given value of x is reversed. For example, when x = -3, the function y = x gives y = (-3) = +9, while the function y = -x gives y = -[(-3) ] = -9. This inverts the graph (see Fig. ) Fig. : Graph of y = -x x y -16 Example 3

12 y = x + 4 This function is obtained by setting a = 1, b = 0 and c = 4, in the general form. Compared with Example 1, we have added a constant, so it should be fairly obvious that this shifts the whole curve bodily upwards by 4 units. The effect of this is that the curve now cuts (intercepts) the y-axis at y = 4. Similarly, if we had subtracted 4 units this would have shifted the whole curve bodily downward. The curve would then cut (intercept) the y-axis at y = -4. Thus we can easily generalise this result and conclude that the parameter c determines the intercept of the curve on the y-axis. Example 4 y = x + 10x Let's consider where this curve will lie in comparison with y = x. When x is positive, so also is 10x, so the graph of y = x + 10x will lie above the graph of y = x. Conversely, when x is negative, so also is 10x, so the graph of y = x + 10x will lie below the graph of y = x. (See Fig. 3). Fig y y = x + 10x y = x x 1 Thus we can see from Fig 3 that the addition of the 10x term has the effect of rotating the graph anticlockwise around the origin.

13 If we had added -10x instead of +10x, obviously the effect would have been reversed, with the graph of y = x - 10x lying below the graph of y = x when x was positive and below it when x was negative. The rotation is thus clockwise in this case. Summary and generalisation Collecting the results of these examples, we can draw the following conclusions about the shape of the quadratic function: y = ax + bx + c (where a, b, and c are parameters) (i) the x term gives the graph a U-shape, called a paraboloa. If the parameter a is positive, the graph looks like a U. If the parameter a is negative, the graph looks like an inverted U (that is, like this: I ). The absolute magnitude of a determines how steeply the curve slopes up (or down). (ii) the constant term, c, determines the intercept of the curve on the y-axis. (iii) the x term causes the parabola to rotate about the graph's y-intercept. If the parameter b is positive, the rotation is anti-clockwise; if negative, clockwise. The absolute magnitude of b determines the strength of the rotational effect.

14 Finally, to test our mental agility we will look at another function that is a close relative of the quadratic function y = ax + bx + c. Consider the function: y = x The question is, what does the graph of this function look like? To answer this, we note that y = x can of course be written as y = x ½ and if we square both sides of this (a step which leaves the relationship between y and x unchanged) we get x = y Thus the function is a quadratic function, but with x rather than y as the dependent variable. Its graph will therefore have the same shape as the graph of y = x would have if it were drawn with y measured along the horizontal axis and x on the vertical (see Fig. 4). Fig. 4 +x +y y = x 3 x = y or y = x ½ y x x -4 -y

15 5 Graphical solution of quadratic equations Fig. 5 Graph of y = x + x y x Recall that, everywhere along the x-axis, y = 0. We see that the graph cuts the x-axis at x = + and -3. Therefore y = 0 at these points. Therefore x + x - 6 = 0 at these points [since we know that y = x + x - 6] Therefore x = + and -3 are the solutions to the quadratic equation x + x + 6 = 0 We can check this algebraically, as follows: Given x + x - 6 = 0. This factorises easily into (x - )(x + 3) = 0 which is true when either x = + or x = -3

16 Graphical solution of simultaneous quadratic equations 14 y 1 10 y = x y = x x Fig. 6 Suppose we are asked to solve the simultaneous equations: y = x - 4 y = x + We could solve them algebraically, by setting the right hand sides of the two equations equal to one another, then solving the resulting (quadratic) equation. But we could also solve them graphically. If we plot the graphs of the two equations, their shapes will be as in Fig. 6 above. Just as we found in Revison Notes with linear simultaneous equations, any point where the two graphs cut one another lies on both, and therefore the co-ordinates of that point must satisfy both equations simultaneously. From the graph, the co-ordinates of the points where the two curves cut are x = -, y = 0 and x = 3, y = 5. These two pairs of values of x and y are therefore the solutions to the simultaneous equations. (Check this for yourself by algebraic methods.)