Such completing the square applying the quadratic formula
Algebra review
Algebra review – A guide for teachers (Years 11–12)
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Answers to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
• Expanding brackets.
• Factorising linear and quadratic expressions.
Motivation
Algebra is the language of mathematics. Just as ordinary language is essential in verbal and written communication, so the ability to read, write and simplify algebraic expres-sions is fundamental to mathematical understanding, problem solving and the commu-nication of mathematical ideas.
The relevant TIMES modules are:
• Algebraic expressions (Year 7)
• Linear equations (Years 7–8)
• Negatives and the index laws in algebra (Years 7–8)• Fractions and the index laws in algebra (Years 8–9)• Special expansions and algebraic fractions (Years 8–9)• Factorisation (Years 9–10)
• Quadratic equations (Years 9–10)
• Indices and logarithms (Years 9–10).Content
Algebraic manipulation
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The index laws were covered in the TIMES modules:
• Negatives and the index laws in algebra (Years 7–8)
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sometimes referred to as binomials. With practice, the middle step above can be left out.
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A guide for teachers – Years 11 and 12 • {9}
Expand and simplify
(x −1)(xn−1+ xn−2+···+ x2+ x +1).
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Factoring
Factoring is the reverse process to expanding. Writing the number 3128 as 8 × 17 × 23 gives us useful information about the number. For example, it tells us that 3128 is divisi-ble by 8 and 17 and 23. It also allows us to calculate 3128÷23 without using a calculator. Similarly, in algebra, factoring an algebraic expression is a very useful and important skill. Indeed, it is very unwise in algebra to expand a complicated collection of brackets unless you have no other option. Factoring is often much the better thing to do. Factor-ing also enables us to solve certain types of equations and will reappear when we look at quadratic equations and again in the module Polynomials.
12xy −15x −16y +20 = 3x(4y −5)−4(4y −5)
= (3x −4)(4y −5).
Quadratics
A quadratic expression has the form ax2+ bx + c, where a,b,c are given numbers with a ̸= 0. In some cases a quadratic expression can be factored. For example, starting with
Clearly the coefficient of x is the sum of a and b and the constant term is their product. This is why we split the 5x term into 2x + 3x. In practice, when factoring x2+ 5x + 6, we seek two integers whose sum is 5 and whose product is 6. Clearly the numbers are 2 and 3, and so we obtain the desired factorisation.
To factor 3x2+14x −5, we first multiply the 3 and the −5 to obtain −15. Now we find two numbers whose sum is 14 and whose product is −15. Clearly, these are 15 and −1. We use these numbers to split the middle term and write
3x2+14x −5 = 3x2+15x − x −5.
Special factorisations
The following three special factorisations are exactly the same as the three special expan-sions discussed in Expanding (Special expansions) in this module, but used in reverse:
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Exercise 5
Algebraic fractions are manipulated and simplified using exactly the same rules that ap-ply to ordinary fractions. In particular, to add or subtract them we need to find (lowest) common denominators. In many instances factoring needs to be done first so that these common factors become apparent.
A guide for teachers – Years 11 and 12 • {13}
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3 (x −1)(x +1). |
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It is now apparent that the lowest common denominator is x(x +1)(x −1). Hence,
For example, to multiply
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x +3 3x2−6x |
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Cancelling out the factors 3x and x +3, we have
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1 (x −3)(x −2). |
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2x −4 = 6 | and |
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are examples of linear equations, whereas
ex= x −3
are often referred to as transcendental equations. With some exceptions, such transcen-dental equations are generally not dealt with in secondary school mathematics.
Linear equations
The various types of linear equations and the various strategies to solve them are dealt with at length in the module Linear equations, and so we will only quickly revise some of these ideas here via two examples. The basic rule throughout is that whatever you do to one side of the equation you must also do to the other.
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Linear inequalities
A linear inequality resembles in form an equation, but with the equal sign replaced by an inequality symbol. The solution of a linear inequality is generally a range of values, rather than one specific value. Such inequalities arise naturally in problems involving words such as ‘at least’ or ‘at most’.
Inequalities also arise when we examine the domain of certain functions. Hence it is important that students are familiar with these before studying functions.
Factoring method
The factoring method relies on the basic fact that if the product of two numbers is zero, then at least one of these numbers must be zero. That is, if AB = 0, then A = 0 or B = 0. Hence, if we have a quadratic equation in which the quadratic has been factored, then we can equate each factor to 0 and solve the resulting linear equation. The method relies on being able to factor the quadratic easily, which is not always possible.
The same procedure works for non-monic quadratics.
We may be given an equation which is a quadratic in disguise, that is, we can rearrange
{18} • Algebra review
Exercise 7
To complete the square on the equation x2−4x −6 = 0, we take the −6 to the other side to produce x2−4x = 6. Now try to write the left-hand side as a perfect square by halving the coefficient of x. The square (x −2)2when expanded gives x2−4x +4, so we add 4 to both sides of the original equation to produce (x −2)2= 6+4 = 10. We can now take the positive and negative square root and solve. Here is the full setting out:
x2−4x −6 = 0
x2−4x = 6
x −2 = | ||
x = 2+ |
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The equation has two irrational solutions.
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The general equation of a circle with centre (a,b) and radius r is (x − a)2+(y −b)2= r2. By completing the square on the terms with x and then the terms with y, find the centre and radius of the circle
x2−2x + y2−6y = 1.
(2ax +b)2= b2−4ac | |||
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2ax +b = |
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or | |
x = −b + |
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{20} • Algebra review
If the quantity b2− 4ac is negative, then there are no solutions. This number is called the discriminant and often denoted by the Greek letter ∆ (pronounced ‘delta’). When solving quadratic equations it is generally a good idea to calculate the discriminant ∆first and think of the quadratic formula as
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This method is often extended to deal with simultaneous equations which are not linear.
Exercise 9
Solve simultaneouslyor subtracting one equation from a multiple of the other.
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A guide for teachers – Years 11 and 12 • {23}
Formulas
For example, the cosine rule states that if a,b,c are the side lengths of a triangle and A is the angle opposite a, then
a2= b2+c2−2bc cos A.
The compound interest formula states that A = P(1+R)n, where P is the principal, R the interest rate per unit time, n the number of units of time, and A the amount to which the principal has increased. Make n the subject of the formula.
History and applications
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Answers to exercises
Exercise 1
A guide for teachers – Years 11 and 12 • {25}
Exercise 5
(a −b)(a +b)3
Exercise 6
−13x
(x +1)(x +2)(x +3)
Exercise 7
x = 2
Exercise 8
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