# Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date

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1 Leaving Certificate Honours Maths - Algebra the date

2 Chapter 1 Algebra This is an important portion of the course. As well as generally accounting for 2 3 questions in examination it is the basis for many other parts of the course. 1.1 Simultaneous Equations Theory Therearethree types of simultaneous equations you need to beable to solve: Two linear equations in x and y One linear and one quadratic equation in x and y Three linear equations in x, y and z This is also important for solving problems with lines and circles in coordinate geometry Exercises Notes: You need to beable to solve the three types of simultaneous equations. Two linear equations in x and y One linear and one quadratic equation in x and y Three linear equations in x, y and z Student Work 1

3 1. LC HL 2009 Paper 1, 2 (a) Solve the simultaneous equations x y +8 = 0 x 2 +xy +8 = 0 2. LC HL 2006 Paper 1, 2 (a) Solve the simultaneous equations 3. LC HL 2003 Paper 1, 2 (a) Solve the simultaneous equations 4. LC HL 2001 Paper 1, 2 (a) Solve the simultaneous equations 5. LC HL 1999 Paper 1, 2 (a) Solve the simultaneous equations 6. LC HL 1997 Paper 1, 2 (a) Solve the simultaneous equations y = 2x 5 x 2 +xy = 2 3x y = 8 x 2 +y 2 = 10 x y = 0 (x+2) 2 +y 2 = 10 x+y = 1 x 2 +y 2 = 25 2x 3y = 1 x 2 +xy 4y 2 = 2 7. LC HL 2007 Paper 1, 2 (a) Solve the simultaneous equations x+y +z = 2 2x+y +z = 3 x 2y +2z = 15 2

4 8. LC HL 2004 Paper 1, 2 (a) Solve, without using a calculator, the following simultaneous equations 3x+y +z = 0 x y +z = 2 2x 3y z = 9 9. LC HL 2002 Paper 1, 2 (a) Solve, without using a calculator, the following simultaneous equations x+2y +4z = 7 x+3y +2z = 1 y +3z = LC HL 2000 Paper 1, 2 (a) Solve for x,y,z 3x y +3z = 1 x+2y 2z = 1 4x y +5z = LC HL 1996 Paper 1, 2 (a) Solve for x,y,z x+y z = LC HL 2005 Paper 1, 1 (a) Solve the simultaneous equations x y +z = 4 x y z = 8 x 5 y 4 = 0 3x+ y 2 = LC HL 1998 Paper 1, 1 (a) Solve for x and y: 2x 5 + y 3 5 = 6 3x 3y 5 +2 =

5 1.2 Quadratic Equations: Solving and Conditions on Roots Theory Solving Quadratic Equations using the Formulae A quadratic equation is an equation of the form f(x) = ax 2 +bx+c = 0 Solvingsuchanequationinvolves findingvaluesofx Rsuchthatf(x) = 0. There are two methods on this course for solving quadratic equations. When the roots of the equation are rational the equation can (in principle) be solved by factorization. Example: Another example: 6x 2 +5x 4 = 0 (2x 1)(3x+4) = 0 2x 1 = 0 and/or 3x+4 = 0 x = 1 2,x = 4 3 6x 2 13x+4 = 0 (2x 1)(3x 4) = 0 etc. If a quadratic equation is given so that the term in front of the x 2 coefficient is negative then it is generally easier to cross multiply the whole equation by 1. i.e. 6x 2 5x+4 = 0 is the same as 6x 2 +5x 4 = 0 etc. Not all quadratics can be solved by this method. For example 2x 2 +5x 4 = 0 Try it. In particular they can t be solved by factorization when the roots are non-rational. Fortunately a more general method exists. Roots α and β can be found for ax 2 +bx+c = 0 a,b,c R using the formulae α,β = b± b 2 4ac 2a 4

6 1 Example 2x 2 +5x 4 = 0. Now a quadratic equation ax 2 +bx+c = 0 can be sketched. When a > 0 it makes a u shape and when a < 0 it makes an n shape. 1 The derivation of this formulae isn t currently on the course, however it may be useful. It works on the principle that quadratic equations can be solved by completing the square. The idea is to make a perfect square involving x on one side of the equation and then solve by getting the square root of both sides. ax 2 +bx+c = 0 x 2 + b a x+ c a = 0 Now: (x+d) 2 = x 2 +2xd+d 2 So comparing equations So d = b 2a and d2 = b2 4a 2 x 2 + b ( ) b 2 a x+ 4a b2 + c 2 4a 2 a = 0 (x 2 ba b2 + x+ ) b2 4a 2 4a + c 2 a = 0 ( x+ b ) 2 b2 2a 4a + c 2 a = 0 ( x+ b ) 2 = b2 2a 4a c 2 a ( x+ b ) 2 = b2 4ac 2a 4a 2 Now taking the square root of both sides and remembering the positive and negative values of the root x+ b 2a = ± b2 4ac 2a Finally the positive and negative values of the square root gives us the two possible roots x = b± b 2 4ac 2a In general all quadratics can be solved by the completing the square method. The quadratic formulae is a systematic application of this technique. 5

7 The Discriminant of a Quadratic Equation The roots of a quadratic equation are specifically the points at which it crosses the x axis. For example has roots at x = 1 and x = 4. x 2 5x+4 = 0 So the quadratic equation ax 2 +bx+c = 0 crosses the x axis at two distinct points has two real roots. Recall our formulae So α,β = b± b 2 4ac 2a α = b+ b 2 4ac and β = b b 2 4ac 2a 2a Now α β so b 2 4ac > 0 i.e. b 2 4ac > 0 Where b 2 4ac is called the discriminant of the equation. If a quadratic equation ax 2 +bx+c = 0 hits the x axis at a single point only, then its two roots are the same (this is called the degenerate case). An example: x 2 4x+4 = 0 has two degenerate roots both at x = 2. In these cases where α = β i.e. the roots are real and equal the discriminant of the quadratic equation b 2 4ac = 0 Now in a quadratic equation ax 2 +bx+c = 0 when b 2 < 4ac this implies that b 2 4ac < 0 6

8 and hence b 2 4ac is trying to find the square root of a negative number and hence the roots of the equation are complex. In this case the graph of the quadratic does not intersect the x axis. For example: 2x = 0. So the discriminant b 2 4ac of the equation ax 2 +bx+c determines the type of roots the equation has: 1. b 2 4ac 0 : real roots 2. b 2 4ac > 0 : distinct real roots 3. b 2 4ac = 0 : equal real roots 4. b 2 4ac perfect square : rational roots (could also solve by factoring) 5. b 2 4ac < 0 : complex roots Difference between Roots of a Quadratic Equation The difference α β between the two roots α,β of a quadratic equation ax 2 +bx+c = 0 can be calculated by subtracting β from α. Now so α = b+ b 2 4ac and β = b b 2 4ac 2a 2a α β = b+ b 2 4ac b b 2 4ac 2a 2a Hence b α β = 2 4ac a This may be useful for solving particular problems with roots Exercises Notes: Solving quadratic equations by formulae and by factorization Conditions on roots of quadratics via discriminant of equation 7

9 Solving inequalities is important here. Solving more complex quadratics via substitution. Student Work 1. LC HL 2009 Paper 1, 2 (b) (a) The graphs of the three quadratic functions f,g and h are shown. In each case, state the nature of the roots of the function. (b) The equation kx 2 +(1 k)x+k = 0 has equal real roots. Find the possible values of k. 2. LC HL 2007 Paper 1, 1 (b) Let f(x) = x 2 +(k+1)x k 2, where k is constant. (a) Find the value of k for which f(x) = 0 has equal roots. (b) Find, in terms of k the roots of f(x) = 0. (c) Find the range of values of k for which both roots are positive. 3. LC HL 2006 Paper 1, 2 (b) Find the range of values of t R for which the quadratic equation (2t 1)x 2 +5tx+2t = 0 has real roots. Explain why the roots are real when t is an integer. 4. LC HL 2004 Paper 1, 2 (c)(ii) Show that for any real values a,b and h, the quadratic equation has real roots (x a)(x b) h 2 = 0 5. LC HL 2003 Paper 1, 1 (c) The real roots of x x + c = 0 differ by 2p where c,p R and p > 0. 8

10 (a) Show that p 2 = 25 c (b) Given that one root is greater than zero and the other root is less than zero find the range of possible values of p. 6. LC HL 2002 Paper 1, 1 (c) (p+r t)x 2 +2rx+(t+r p) = 0 is a quadratic equation, where p,r and t are integers. Show that (a) the roots are rational (b) one of the roots is an integer 7. LC HL 2002 Paper 1, 2 (c) (a) Showthat iftherootsofx 2 +bx+c = 0differby1thenb 2 4c = 1. (b) The roots of the equation x 2 +(4k 5)x+k = 0 are consecutive integers. Find the values of k and the roots of the equation. 8. LC HL 2000 Paper 1, 2 (b) Solve x 2 2x 24 = 0. Hence find the values of x for which ( x+ x) 4 2 ( 2 x+ 4 ) 24 = 0 x x R,x LC HL 1998 Paper 1, 1 (c) If the quadratic equation ax 2 +bx+c = 0 has equal roots, solve for x in terms of a and b where a,b R. Now by letting x = 3 y write t3 y +3 y = 3 as a quadratic equation in x, where t R and t 0. Find the value of t for which this equation has equal roots. Assuming this value of t solve the equation t3 y +3 y = LC HL 1998 Paper 1, 2 (c) Solve x 2 6x+8 = 0 and hence find the values of x for which ( 1+ x) 1 2 ( ) +8 = 0 x x R,x 0 9

11 1.3 Solving Inequalities Theory Quadratic Inequalities You may be asked to solve quadratic inequalities, i.e. Find a range of values of x such that x 2 +3x 4 0 The first thing to do with this quadratic inequality is to observe the shape of the graph. If the coefficient in front of the x 2 term (a) is greater than zero then the graph is a u shape if negative then it is an n shape. Sketch the graph, the points where the graph crosses the x axis are the roots of the quadratic equation, i.e. in our example x 2 +3x 4 = 0 Solve the equation and find the roots (either by factoring or using the formulae). x = 4 and x = 1 Now mark these points in or your sketch. Now, the portion of the graph that is below the x axis represents the range of values where cx 2 +3x 4 0 These are the values { 4 x 1} Similarly the portion of the graph that is above the x axis represents the range of values of x where x 2 +3x

12 These are the values Thus giving us the required result. {x 4} {x 1} Absolute Values Inequality questions are also asked using absolute values i.e. 2x 3 < 4 These are really quadratic inequalities. The problem with these is that what the question is really asking is find values of x such that 2x 4 < 4 and 2x 4 < 4 Thesecan betricky to keep track of. A better way todothis typeof problem is to use the definition of the absolute value a = a 2 to form (2x 3) 2 < 4 Squaring both sides Then (2x 3) 2 < 16 4x 2 12x+9 < 16 Finally forming a quadratic inequality which we can solve 4x 2 12x 7 < 0 Inequalities with Fractions Inequalities with fractions generally look something like: Find the range of values for which 2x 3 x+2 1 Now, theproblemwiththeseisthatunless we are explicitly told that x+2 > 0 that we cannot simply cross multiply across the inequality as we do not know if x + 2 is positive or negative. This then results in having to deal with positive and negative cases. A better way is to multiply the numerator and the denominator of the fraction by (x+2). (2x 3)(x+2) (x+2)

13 As (x + 2) 2 > 0 2 we can then multiply this safely across the inequality without worrying about the sign. (2x 3)(x+2) (x+2) 2 Now 2x 2 +x 6 x 2 +4x+4 Which resolves itself to a quadratic inequality which we can solve x 2 3x 10 0 Proofs with Inequalities These can be difficult. Generally they begin with some inequality and then you are asked to prove that it is true. The method of proof generally revolves around taking all of the values to one side of the inequality (using the methods previously described) so you have some terms 0 Then if you can prove that the left hand side of the inequality is a perfect square the proof is complete as a square is always positive Exercises Notes: x+1 < 3 can be solved by (x+1) 2 < 3 (x+1) 2 < 9 x 2 +2x+1 9 < 0 x 2 +2x 8 < 0 and then solve the resulting quadratic to find values for x Proving identities with inequalities generally involve proving that one side can be made into a perfect square and hence is greater (or equal to) zero Only positive values 0 can be multiplied across an inequality without changing the sign. 2 All perfect squares are positive. Try it! 12

14 Multiplying the denominator of a fraction by itself will ensure that the denominator > 0 and can hence be multiplied across an inequality without changing the sign. Student Work 1. LC HL 2008 Paper 1, 5 (a) Find the range of values of x that satisfy the inequality x 2 3x LC HL 2007 Paper 1, 5 (a) Plot, on the number line, the values of x that satisfy the inequality x Z x LC HL 2005 Paper 1, 2 (a) Solve for x: x 1 < 7, where x R 4. LC HL 2004 Paper 1, 2 (b)(i) Solve the inequality x+1 < 4 where x R and x 1. x 1 5. LC HL 2003 Paper 1, 2 (b)(i) Solve for x: 4x+7 < 1 6. LC HL 2002 Paper 1, 2 (b)(i) Find the range of values of x R for which x 2 +x LC HL 2001 Paper 1, 2 (b)(i) Solve for x: 3x+5 < 4 8. LC HL 1998 Paper 1, 1 (a) Solve for x x 4 < 5 9. LC HL 1997 Paper 1, (b) (ii) Solve x+3 x 4 < 2 x 4 and x R. 13

15 10. LC HL 1996 Paper 1, 2 (b)(i) Solve for x 2x 7 x+3 < 1 x LC HL 2009 Paper 1, 1 (b) Let f(x) = x 2 7x+12. (a) Show that if f(x+1) 0, then f(x) f(x+1) simplifies to x 4 (b) Find the range of values of x for which f(x) f(x+1) > LC HL 1999 Paper 1, 1 (b) Solve for x: x R, x 3. 4x 1 x 3 < 2 x LC HL 2008 Paper 1, 2 (c) Show that if a and b are non-zero real numbers, then the value of a b + b a can never lie between 2 and 2. [Hint: consider the case where a and b have the same signs separately from the case where a and b have opposite signs] 14. LC HL 2007 Paper 1, 2 (c) (a) Prove that x+ 9 x+2 (b) Prove that x+ 9 x+a 4, where x+2 > 0. 6 a, where x+a > LC HL 1998 Paper 1, 2 (c) Show that for all real numbers a,b > 0 then a+b 2 ab Show that Hence, deduce that (a+b)( 1 a + 1 b ) = (a+b)2 ab (a+b)( 1 a + 1 b ) 4 14

16 16. LC HL 1996 Paper 1, 1 (c)(i) Make a sketch of the region of the plane represented by y x and y 2+ x 17. LC HL 2005 Paper 1, 5 (c) Show that a+b a 2 +b where a and b are real numbers. Now the lengths of a right angled triangle are a, b and c where c is the length of the hypotenuse. Show that a+b c LC HL 1999 Paper 1, 5 (c) Leta,b,cbeunequal, realpositivenumbers. Usingtheresultsa 2 +b 2 > 2ab, b 2 +c 2 > 2bc and c 2 +a 2 > 2ac deduce: (a) a 2 ab+b 2 > ab (b) a 2 +b 2 +c 2 > bc+ca+ab (c) a 2 +b 2 > ab(a+b) 1.4 Functions, Factorization and Simplification Theory Functions When substituting a value a into an function f(x) =..., everywhere there is an x in the right hand term of the expression substitute in the value for a. This also works if a is itself a function, fraction etc. A method that may prove useful in understanding these questions is to change the placeholder variable from x to y i.e. g(y) =... etc. Factorization As well as the factors for x 2 y 2 = (x+y)(x y) x 3 y 3 = (x y)(x 2 +xy +y 2 ) x 3 +y 3 = (x+y)(x 2 xy +y 2 ) it is also useful to know the factorization for x 4 y 4 15

17 That is x 4 y 4 = (x 2 ) 2 (y 2 ) 2 which is the difference of two squares and can be factored in the usual way (x 2 ) 2 (y 2 ) 2 = (x 2 +y 2 )(x 2 y 2 ) The second factor here is the difference of two squares again, so Exercises Notes: x 4 y 4 = (x 2 +y 2 )(x+y)(x y) When substituting in values in to the function f(x) if the values are abstract it may be useful to relabel the holder variable to y instead of x i.e. f(y) =... and go from there. x 4 y 4 = (x 2 +y 2 )(x+y)(x y) Student Work 1. LC HL 2009 Paper 1, 1 (a) Find the value of x y when 2x+3y x+6y = LC HL 2008 Paper 1, 1 (a) Simplify fully x 2 +4 x 2 4 x x+2 3. LC HL 2008 Paper 1, 2 (a) Express x 2 +10x+32 in the form (x+a) 2 +b. 4. LC HL 2007 Paper 1, 1 (a) Simplify x 2 xy x 2 y 2 5. LC HL 2005 Paper 1, 1 (b)(i) Express in the form 2 p q, where p,q Z. 16

18 6. LC HL 2004 Paper 1, 1 (b)(ii) Show that 3 1+x p x p simplifies to a constant. 7. LC HL 2003 Paper 1, 1 (a) Express the following as a single fraction in its simplest form 6y x(x+4y) 3 2x 8. LC HL 2001 Paper 1, 1 (a) Find the real numbers a and b such that for all x R. x 2 +4x 6 = (x+a) 2 +b 9. LC HL 2000 Paper 1, 1 (a) Show that the following simplifies to a constant when x 2 3x 5 x x 10. LC HL 2002 Paper 1, 2 (b)(ii) Let g(x) = x n +3 for all x R where n N. Show that if n is odd then g(x)+g( x) is a constant. 11. LC HL 2001 Paper 1, 1 (b)(ii) Simplify (x x2 )(x x2 ) andexpressyouranswerintheformx n + 1 x n wherenisawholenumber. 12. LC HL 2005 Paper 1, 2 (c) Let f(x) = x2 +k 2, where k and m are constants and m 0. mx (a) Show that f(km) = f( k m ) (b) a and b are real numbers such that a 0, b 0 and a b. Show that if f(a) = f(b) then ab = k LC HL 2004 Paper 1, 1 (c) Show that p 3 +q 3 (p+q) 3 = 3pq(p+q) Then find in terms of a and b the three values of x for which (a x) 3 +(b x) 3 (a+b 2x) 3 = 0 17

19 14. LC HL 2004 Paper 1, 2 (c)(ii) f(x) = 2x+1 for all x R. Show that there exists a number k such that for all x, f(x+f(x)) = kf(x). 15. LC HL 2000 Paper 1, 2 (c) Express a 4 b 4 as the product of three factors. Now factorize Using these results show that a 5 a 4 b ab 4 +b 5 a 5 +b 5 > a 4 b+ab LC HL 1997 Paper 1, 2 (c) Let f(x) = 1 x for all x R and x 0. Points a and b have coordinates (p,f(p)) and (q,f(q)) respectively for 0 < p < q. (a) Show that the equation of the line ab can be written as (b) Show that y = g(x) = 1 p 1 pq (x p) f(x) g(x) = (x q)(x p) pqx Hence show that f(x) g(x) < 0 for 0 < p < x < q. 1.5 Square Roots(Surds) Logs and Exponentials Theory Index Notation a 2 = a a a 1 2 = a a 1 = 1 a a 3 2 = ( a) 3 = a a a 1 2 = 1 a a b a c = a b+c 18

20 ( ) a c b = a c b c (a b ) c = a bc a 0 = 1 Square Roots (Surds) ab = a b a b = a b a a = a Rationalizing the Denominator: a b = a b b b = a b b Rationalizing the Denominator: a b+ c = ( a b+ c Rationalizing the Denominator: )( b c b c ) ( )( a = a b+ c b c b c Solving Equations Involving Surds Equations involving surds generally occur in two types: a = b = a(b c) b 2 c b+ c ) = a( b+ c) where a and b are some functions of a variable (generally x). This can be solved by squaring both sides. so ( a) 2 = b 2 a = b 2 which will generally resolve to a quadratic equation in x which can be solved by the normal methods. The second type of equation that must be solved is of the form a = b+c b c wherea, bandcarefunctionsofx. Thesecanbesolvedasbeforebysquaring both sides ( a) 2 = ( b+c) 2 a = ( b) 2 +2c b+c 2 a = b+2c b+c 2 19

21 Notice that we are still left with a b term, which must be removed. To do this bring all terms without surds to one side of the equation and square again i.e. (a b c 2 ) 2 = (2c b) 2 This results in an equation (generally a quadratic) of the form (a b c 2 ) 2 = 4c 2 b where a,b,c are functions of x which in principle can be solved. Note: It is very important to check back the results of these equations! Generally negative answers are disallowed if results must be real. (Square root of a negative number etc.). Sometimes when squaring an extra root is introduced which does not in fact satisfy the equation this is called the extraneous root. Logarithms A log is the power a number must be raised to to give another number i.e. two to the power of four is sixteen so the log to the base two of sixteen is four. In general if b a = c then log b c = a Now log b a+log b c = log b (ac) ( log b a log b c = log a ) b c log b (a c ) = clog b a log b b = 1 log b 1 = 0 An important fact about logs is the idea of a change of base. Now if c = a then we get log b a = log ca log c b log b a = log aa log a b = 1 log a b This is fact is useful for some logarithmic identities. In general when solving problems with logs the first thing to do is to bring all logs in the problem to the same base. This can be achieved using the above identities. Note that log0 and the log of a negative numbers does not exist. So in general for log(a) a 0. Solving Equations Involving Logs The approach to equations with logs is to try to: 20

22 Get all logs to the same base using the change of base rule Then if there are logs and numbers in the equation bring all the logs to one side and all of the numbers to the other. Now using the definition of a log log b a = c b c = a form an equation (usually a quadratic) and solve it in the normal manner. If there are logs to the same base on both sides of the equation, then raise both sides of the equation to the power of the base to remove the logs i.e log b a = log b c a = c Then from here solve the equation in the normal way. (Generally a and c are functions of some variable usually x.) Note that some of these (quadratic) equations will result in extraneous roots as a result of the exponentiations, so as for equations involving surds the solutions will need to be checked for correctly. Any negative solutions or solutions equal to zero can be immediately discarded Exercises Notes: Indices: (Work these out both ways to ensure that they make sense to you) 4 2 = 4 4 = = 4 = = = ( 4) 3 = 4 4 = 4 2 = = 1 4 = = = 2 5 = 32 (2 3 ) 2 = = 2 6 = 64 Surds (Square Roots): 3 3 = 3 6 = 2 3 = =

23 Logs: Rationalizing the Denominator: 2 3 = = 2 3 ( 2 Rationalizing the Denominator: 3+ = = ( 2 Rationalizing the Denominator: 5 3 = 2( 5+ 3) 5 3 = 5+ 3 log 3 27+log 3 9 = log 3 (243) ( log 2 27 log 3 9 = log 27 ) b 9 = log3 3 = 1 log 3 (9 2 ) = 2log 3 9 = 4 log 3 3 = 1 log 3 1 = 0 log 3 9 = log 29 log 2 3 log 3 9 = log 33 log 9 3 = 1 log 9 3 = 1 1 = 3 3 Student Work 1. LC HL 1999 Paper 1, 1 (a) Show that = )( ) )( = 2(3 5) = 5+ 3 ) = 2. LC HL 1997 Paper 1, 1 (a) If x = 1+ 1 a and y = a 1 a, a > 0, find the value of x 2 y LC HL 1996 Paper 1, 1 (a) Express in the form a 2 b where a,b N 4. LC HL 2006 Paper 1, 1 (a) Find the real number a such that for all x 9 x 9 x 3 = x+a 5. LC HL 2009 Paper 1, 5 (a) Solve for x: x 2 = 3x 2 22

24 6. LC HL 2002 Paper 1, 2 (a) Solve the equation x = x+2 7. LC HL 2005 Paper 1, 5 (a) Solve for x: 10 x = 4 x 8. LC HL 2003 Paper 1, 5 (a) Solve for x: x = 7x LC HL 1999 Paper 1, 5 (b)(i) Solve the equation 2x+7 = 2+ x 10. LC HL 2002 Paper 1, 5 (a) Find the value of x in each case 8 (a) 2 = 32 x (b) log 9 x = LC HL 1997 Paper 1, 5 (a) Solve ( ) 3 log 5 x = 1+log 5,x R,x > 1 2x LC HL 1996 Paper 1, 5 (a) Solve the simultaneous equations where x > 1,y > LC HL 2008 Paper 1, 5 (b) Solve the equations log(x+y) = 2log x log y = log 2+log(x 1) (a) 2 x2 = 8 2x+9 (b) log e (2x+3)+log e (x 2) = 2log e (x+4) 14. LC HL 1998 Paper 1, 5 (b) Solve log 5 (x 2) = 1 log 5 (x 6) where x R,x > 6 23

25 15. LC HL 2004 Paper 1, 5 (b)(ii) Solve log 4 x log 4 (x 2) = LC HL 2001 Paper 1, 5 (b)(i) Solve log 6 (x+5) = 2 log 6 x for x LC HL 2000 Paper 1, 5 (c) Solve for x (a) 2log 9 x = 1 2 +log 9(5x+18),x > 0 (b) 3e x 7+2e x = LC HL 1999 Paper 1, 5 (b)(ii) If x > 0 and x 1, show that Note: 1 log 2 x + 1 log 3 x + 1 log 5 x = 1 log 30 x log b a = log ca log c b 19. LC HL 2006 Paper 1, 5 (c) Given two real numbers a and b where a > 1 and b > 1, prove that Under what conditions is 1 log b a + 1 log a b 2 1 log b a + 1 log a b = LC HL 2005 Paper 1, 5 (c) Show that 1 log a b = log ba where a,b > 0 and a,b = 1. Hence show that where c > 0, c 1. 1 log 2 c + 1 log 3 c + 1 log 4 c log r c = 1 log r! c 24

26 21. LC HL 2009 Paper 1, 5 (c) Solve the simultaneous equations: log 3 x+log 3 y = 2 log 3 (2y 3) 2log 9 x = LC HL 2006 Paper 1, 2 (c) f(x) = 1 b 2x and g(x) = b 1+2x, where b is a positive real number. Find in terms of b the value of x for which f(x) = g(x). 23. LC HL 2003 Paper 1, 2 (c)(i) Solve for y: 2 2y+1 5(2 y )+2 = The Factor Theorem: Factoring and Solving Cubic Equations Theory The roots of a function f(x) are the values k such that f(k) = 0. In this course we are mainly concerned with quadratic and cubic ax 2 +bx+c ax 3 +bx 2 +cx+d functions. The roots of a function are the values of x where the graph of the function crosses the x axis. A quadratic function can have two, one (degenerate) or zero real roots. In the case of zero real roots both roots are complex (as previously shown). A cubic equation has three roots. One root must be real and the other two are either then both complex or both real. The Factor Theorem for Quadratics Let f(x) be a quadratic such that f(x) = ax 2 +bx+c Now for k R,f(k) = 0 f(k) = ak 2 +kx+k = 0 so f(x) f(k) = a(x 2 k 2 )+b(x k) = a(x k)(x+k)+b(x k) = (x k)[a(x+k)] 25

27 So, (x k) is a factor of f(x) f(k), now if k is a root of f(x) i.e. f(k) = 0 we can see that (x k) is a factor of f(x). The Factor Theorem for Cubics Let f(x) be a cubic such that f(x) = ax 3 +cx 2 +cx+d Now for k R, f(k) = 0 so now so f(k) = ak 3 +bk 2 +ck +d = 0 f(x) f(k) = a(x 3 k 3 )+b(x 2 k 2 )+c(x k) = a(x k)(x 2 +xk+k 2 )+b(x+k)(x k)+c(x k) = (x k)[a(x 2 xy +y 2 )+b(x+k)+c] Thus in general (x k) is a factor of f(x) f(k). Now if f(k) = 0 i.e. k is a root of f(x) then (x k) is a factor of f(x) Solving Problems using the Factor Theorem In general if k is a root of f(x) then (x k) is a factor. This gives us a convenient way to solve cubic equations. For example if we are told that x = 1 2 is a root of 2x 3 +7x 2 +7x+2 = 0 then we can find the other two roots as follows: x = 1 2 is a root. Hence x = 0 and multiplying across by 2 we get a factor of the quadratic (2x+1) Now (2x+1) is a factor of 2x 3 +7x 2 +7x+2 so this means it divides into it without a remainder! try it!. Using old fashioned long division we see that the other factor of the is x 2 +3x+2 so 2x 3 +7x 2 +7x+2 = (2x+1)(x 2 +3x+2) = 0 Now the quadratic x 2 +3x+2 = 0 can be solved in the normal manner to find the other two roots. 26

28 If you are given a cubic without any roots and told to solve it 3 : You musttry tofindthefirstroot bytrial and error bytesting values k until oneof then makes f(k) = 0. Normally (the typeof questions that are asked) it is sufficient to test the values: k = {1, 1,2, 2,3, 3} and one of these should yield a result. If one of the roots of ax 3 +bx 2 +cx+d is an integer then it is a factor of d a 4. So if a = 1 then the root is a factor of d. Then form a factor from this root using the factor theorem. Long divide this factor in to the cubic, and you will be left with a quadratic equation. Solve/Factor this quadratic by the usual methods to get the other two roots Exercises Notes: Difference of two squares: x 2 y 2 = (x y)(x+y) Difference of two cubes: x 3 y 3 = (x y)(x 2 +xy +y 2 ) If k is a root of f(x) = ax 3 +bx 2 +cx+d then f(k) = 0 and (x k) is a factor of ax 3 +bx 2 +cx+d If one polynomial is a factor of another it long divides into it without any remainder. 3 A variant on this method for cubics: If a cubic is written in the form such that its leading term is 1 i.e. x 3 +ax 2 +bx+c and k is a root, i.e. (x k) is a factor then the quadratic factor can be found by inspection without long division via a x 3 +ax 2 +bx+c = (x k)(x 2 +(a+k)x+(b+ak +k 2 )) For example: 1 is a root of x 3 6x 2 +11x 6 so x 3 6x 2 +11x 6 = (x 1)(x 2 +( 6+1)x+(11 6+1)) x 3 6x 2 +11x 6 = (x 1)(x 2 5x+6) The quadratic can then be factored/solved in the usual way 4 see next section for proof. 27

29 Solving a cubic: find one root by trial and error and then use the factor theorem to turn this into a factor of the cubic. Long divide this into the expression to find the other cubic factor. Find the two roots of the quadratic factor using the normal methods. Student Work 1. LC HL 2008 Paper 1, 1 (b) Given that one of the roots is an integer, solve the equation 6x 3 29x 2 +36x 9 = 0 2. LC HL 2005 Paper 1, 1 (b)(ii) Letf(x) = ax 3 +bx 2 +cx+d. Showthat (x t)is afactor of f(x) f(t) 3. LC HL 2005 Paper 1, 2 (b) Thecubicequation 4x 3 +10x 2 7x 3 = 0has oneinteger root andtwo irrational roots. Express the irrational roots in their simplest form. 4. LC HL 2004 Paper 1, 1 (b)(i) Let f(x) = x 3 + kx 2 4x 12, where k is a constant. Given that (x+3) is a factor of f(x), find the value of k. 5. LC HL 2002 Paper 1, 1 (b) The cubic equation x 3 4x 2 +9x 10 = 0 has one integer root and two complex roots. Find the three roots. 6. LC HL 2000 Paper 1, 1 (b) f(x) = ax 3 +bx 2 +cx +d where a,b,c,d R. If k is a real number such that f(k) = 0, prove that (x k) is a factor of f(x) 7. LC HL 1998 Paper 1, 1 (b) If (2x 1) is a factor of the polynomial P(x) = 2x 3 5x 2 kx+3 find the value of k. Then find the other two factors of P(x). 8. LC HL 2006 Paper 1, 1 (b) f(x) = 3x 3 +mx 2 17x+n, where m and n are constants. Given that x 3 and x+2 are factors of f(x), find the value of m and the value of n. 9. LC HL 2001 Paper 1, 1 (b) Let f(x) = 2x 3 +mx 2 +nx+2 where m and n are constants. Given that x 1 and x+2 are factors of f(x), find the value of m and the value of n. 28

30 10. LC HL 1996 Paper 1, 1 (b)(i) (x+1) is a factor of x 3 +5x 2 +kx 12. Find the value of k and the other two factors of the cubic expression. 11. LC HL 1997 Paper 1, 1 (c) If (x 1) 2 is a factor of ax 3 +bx+1 find the value of a and the value of b 1.7 The Factor Theorem: Problems with Quadratics and Cubics Theory Other problems involving roots of quadratic, cubic expressions (polynomials) and the factor theorem. Generally they can be solved by one of two methods. Long division of polynomials or equating coefficients using the factor theorem. Both methods are discussed here. Solving Problems using the Factor Theorem when one factor is given If one factor is given i.e. (x γ) and some function f(x) (which may be quadratic or cubic) and you are asked to prove some identities in terms of f(x) a good strategy is to use the factor theorem to obtain the root and then form the equation x = γ f(γ) = 0 Now as this is equal to zero each of the coefficients must separately equal zero, generally proving the identity. Alternately this type of problem can be solved by long dividing (x γ) into f(x) and forming relationships using the idea that (x γ) cannot leave remainder. In general there are two methods of solving the more difficult factor theorem problems 1. Use the factor to form roots, and then create an equations by substituting roots into the polynomial and setting equal to zero. 2. Use the factor theorem to create factors from roots, then long divide these factors into the polynomial. The factor must divide in equally. This allows the inferring of relationships in the coefficients. 29

31 Relationship between Roots and Coefficients of a Cubic Equation Consider the cubic equation ax 3 +bx 2 +cx+d = 0 Providing a = 0 we can divide across by a to get ( ) b ( ( ) c d x 3 + x 2 + x+ = 0 a a) a and let its three roots be α, β and γ. Now via the Factor Theorem we know that the factors of the the cubic ( ) b ( ( ) c d x 3 + x 2 + x+ a a) a are (x α), (x β) and (x γ) respectively. So x 3 + ( ) b a x 2 + ( ( c a) x+ d ) a = (x γ)(x β)(x α) = (x γ)(x 2 (α+β)x+(αβ)) = x 3 (α+β +γ)x 2 +(γ(α+β)+(αβ)x+(αβγ) So equating coefficients of powers of x on both sides of the equation x 2 : b = (α+β +γ) a x 1 : c a = γ(α+β)+αβ x 0 = 1 : d a = αβγ This derivation is not strictly a required proof on the course but knowledge if it offers a useful method of solving many problems with cubic equations without having to perform long division on polynomials. Solving Problems using the Factor Theorem when quadratic factor is given Generally theseproblemsinvolve beinggiven afactor ax 2 +bx+c(sometimes the quadratic factor can be given as a perfect square (x α) 2 ) of a cubic expression a x 3 +b x 2 +c x+d. If you can find two unique roots α,β easily it is better to get these directly and then use the factor theorem to solve whatever you need in the cubic by f(α) = 0, f(β) = 0. More interesting is the case where you cannot easily get the roots of ax 2 +bx+c. In this case there are two ways to progress Long divide ax 2 +bx+c into a x 3 +b x 2 +c x+d such that there is no remainder. 30

32 Let x = γ be the third root. Here (x γ) is hence a factor so (x γ)(ax 2 +bx+c) = a x 3 +b x 2 +c x+d Now multiplying out and equating coefficients of x 3, x 2, x 1 gives potentially 4 equations and a convenient mechanism of finding all three roots or proving any required relationship Exercises Notes: Try to substitute roots into expressions first if they are readily available The long division without remainder approach is always available to you. If necessary introduce the third root γ and hence factor (x γ). Equating coefficients of expressions is generally easier than the long division approach Perfect squares give a free root. Student Work 1. LC HL 2009 Paper 1, 1 (c) Given that x c + 1 is a factor of x 2 5x + 5cx 6b 2, express c in terms of b. 2. LC HL 2008 Paper 1, 1 (c) Two of the roots of the equation ax 3 +bx 2 +cx+d = 0 are p and p. Show that bc = ad 3. LC HL 2007 Paper 1, 1 (c) x+p is a factor of both ax 2 +b and ax 2 +bx ac. (a) Show that p 2 = b b ac a and that p = b. (b) Hence show that p 2 +p 3 = c 4. LC HL 2006 Paper 1, 1 (c) x 2 t is a factor of x 3 px 2 qx+r. (a) Show that pq = r. (b) Express the roots of x 3 px 2 qx+r = 0 in terms of p and q. 31

33 5. LC HL 2005 Paper 1, 1 (c) (x p) 2 is a factor of x 3 +qx+r. Show that 27r 2 +4q 3 = 0. Express the roots of 3x 2 +q = 0 in terms of p. 6. LC HL 2003 Paper 1, 1 (b) f(x) = ax 2 +bx+c where a,b,c R. Given that k is a real number such that f(k) = 0 prove that x k is a factor of f(x). Now, show that 2x 3 is a factor of 4x 2 2(1+ 3)x+ 3 and find the other factor. 7. LC HL 2003 Paper 1, 2 (b)(ii) Given that x 2 ax 3 is a factor of x 3 5x 2 +bx+9 where a,b R, find the value of a and the value of b. 8. LC HL 2001 Paper 1, 1 (c) x 2 px+q is a factor of x 3 +3px 2 +3qx+r. (a) Show that q = 2p 2. (b) Show that r = 8p 3. (c) Find the three roots of x 3 +3px 2 +3qx+r = 0 in terms of p. 9. LC HL 2000 Paper 1, 1 (c) (x t) 2 is a factor of x 3 +3px+c. Show that (a) p = t 2 (b) c = 2t LC HL 1999 Paper 1, 1 (c) x 2 +bx+c is a factor of x 3 p. Show that (a) b 3 = p (b) c 3 = p LC HL 1996 Paper 1, 1 (c)(ii) x 2 px+1 is a factor of ax 3 +bx+c where a 0. Show c 2 = a(a b) 1.8 Quadratic Equations: Formulae for roots Theory Certain problems involving quadratics involve expressing the roots α and β in terms of the coefficients a, b and c of the quadratic ax 2 + bx + c = 0. If this method can be used it generally yields easier solutions to problems than finding the roots directly. 32

34 The Roots of a Quadratic Equation in terms of its coefficients Consider the quadratic equation ax 2 +bx+c = 0 We know that the roots α and β can always be found using the formulae where α,β = b± b 2 4ac 2a α = b+ b 2 4ac and β = b b 2 4ac 2a 2a Now we want to get some easier relationship between the roots α and β of the quadratic and the coefficients a,b and c than this. First assume a 0 and divide across ax 2 +bx+c = 0 by a to get ( ) b ( c x 2 +x + = 0 a a) Add the roots α and β α+β = ( ) b+ b 2 4ac 2a +( b b 2 4ac 2a b+ b = 2 4ac b b 2 4ac 2a 2b = 2a = b a So we see that ( b a ) is equal to (α+β) the negative of sum of the roots of the quadratic equation. Now multiply the roots α and β ( )( ) αβ = b+ b 2 4ac b b 2 4ac 2a 2a = b2 +b b 2 4ac b b 2 4ac (b 2 4ac) 4a 4ac 2 = 4a c 2 = a So we see that ( c a ) is equal to (αβ) the product of the roots of the quadratic equation. This gives us the relationship for a quadratic equation ) with roots α and β that x 2 +x( b a )+(c a ) = 0 x 2 x(α+β)+(αβ) = 0 33

35 where the sum of the roots α+β = b a and the product of the roots αβ = c a In general for a quadratic equation the following relationship holds x 2 x(sum of roots)+(product of roots) = 0 This is very useful for solving problems related to quadratic equations without ever explicitly finding the roots. Alternative derivation of this relationship Consider a quadratic equation ax 2 +bx+c = 0 where a 0. As before dividing across by a gives ( ) b ( c x 2 +x + = 0 a a) Now let α and β be the roots of this equation. The factor theorem ensures that the factors are (x α) and (x β), so the following relationship holds ( ) b ( c x 2 +x + = (x α)(x β) = x a a) 2 x(α+β)+(αβ) Now this means that the coefficients of x 2, x and the constant term must be equal on both sides of the equation. Looking at the coefficients of x we see that (α+β) = b a and looking at the constant term we see that as before. 5 (αβ) = c a 5 It may be useful to observe that in certain circumstances the relationships α+β = c 1 (where c 1 = b ) and αβ = a c2 (where c2 = c ) can be used to find the two roots α and β of a the quadratic by solving the simultaneous equations and α+β = c 1 αβ = c 2. Observe that this, unless explicitly asked, is probably not simpler than solving the original quadratic ax 2 +bx+c = 0. 34

36 Quadratic Equations without Explicitly Solving There are relationships on roots α and β of a quadratic equation as follows: ax 2 +bx+c = 0 α+β = b a and αβ = c a Observe that in general a quadratic equation can be formed from its two roots as follows x 2 x(sum of roots)+(product of roots) = 0 So if roots of a quadratic are α and β and α+β and α,β are known we can form a number of quadratic equations with roots: Roots: α 2 and β 2 The sum of the roots is α 2 +β 2 Now (α+β) and (αβ) are known so, The product of the roots is Roots: α 2 β and β 2 α The sum of the roots is The product of the roots is Roots: 1 α 2 and 1 β 2 The sum of the roots is The product of the roots is (α+β) 2 = α 2 +2αβ +β 2 α 2 +β 2 = (α+β) 2 2(αbeta) α 2 β 2 = (αβ) 2 α 2 β +β 2 α = (αβ)(α +β) (α 2 β)(β 2 α) = (αβ) 3 1 α β 2 = α+β α 2 β 2 = (α+β) (αβ) 2 1 α 2 1 β 2 = 1 (αβ) 2 35

37 Roots: α 3 and β 3 The sum of the roots is α 3 +β 3 = (α+β)(α 2 αβ +β 2 ) = (α+β)(α 2 +β 2 αβ) But α 2 +β 2 = (α+β) 2 2αβ), so The product of the roots is Roots: 1 α 3 and 1 β 3 The sum of the roots is α 3 +β 3 = (α+β)((α+β) 2 3(αβ)) α 3 β 3 = (αβ) 3 1 α β 3 = α3 +β 3 α 3 β 3 = (α+β)((α+β)2 3(αβ)) (αβ) 3 The product of the roots is Exercises Notes: ax 2 +bx+c = 0 x 2 x(α+β)+(αβ) = 0 α+β = b a αβ = c a α 2 +β 2 = (α+β) 2 2(αβ) 1 1 α 3 β 3 = 1 (αβ) 3 α 3 +β 3 = (α+β)(α 2 2αβ +β 2 ) Student Work 1. LC HL 2009 Paper 1, 2 (c) (a) One of the roots of px 2 + qx + r = 0 is n times the other root. Express r in terms of p,q and n (b) One of the roots of x 2 +qx+r = 0 is five times the other. If q and r are positive integers, determine the set of possible values of q 36

38 2. LC HL 2008 Paper 1, 2 (b) α and β are roots of the equation x 2 7x+1 = 0. (a) Find the value of α 2 +β 2 (b) Find the value of 1 α β 3 3. LC HL 2007 Paper 1, 2 (b) α and β are the roots of the equation x 2 4x+6 = 0. (a) Find the value of 1 α + 1 β (b) Find the quadratic equation whose roots are 1 α and 1 β. 4. LC HL 2004 Paper 1, 2 (b)(ii) The roots of x 2 +px+q = 0 are α and β, where p,q R. Find the quadratic equation whose roots are α 2 β and αβ LC HL 2003 Paper 1, 2 (c)(ii) Given that x = α and x = β are roots of the quadratic equation 2k 2 x 2 +2ktx+t 2 3k 2 = 0 where k,t R and k 0, show that α 2 +β 2 is independent of k and t. 6. LC HL 2001 Paper 1, 2 (c) α and β are real numbers such that α+β = 7 and αβ = 11. (a) Show that α 2 +β 2 = 27. (b) Find a quadratic equation with roots α β and β α and write your answer in the form px 2 +qx+r = 0 where p,q,r Z. 7. LC HL 1998 Paper 1, 2 (b) If α and β are roots of the equation x 2 6x+2 = 0 α > 0, β > 0, find αβ and α + β. Now factorize α 3 + β 3. Find the value of α 3 +β LC HL 1996 Paper 1, 2 (c) Find the quadratic equation with roots 1 α and 1 β given that α+β = 5 and αβ = k where k 0. Now find the range of values of k for which the equation will have real roots. 37

39 1.9 Summary To fully answer algebra questions you need to be able to do the following: Solve simultaneous equations two linear equations in two variables three linear equations in three variables a quadratic equation and a linear equation in two variables Recognize and solve quadratic equations by factorization and using the formulae. Recognize and solve more complex quadratic equations by substitution. Get the discriminant of a quadratic equation and use the value of the discriminant to determine conditions on the roots. Recognize and solve linear and quadratic inequalities. Convert absolute values to quadratic inequalities and solve. Prove identities with inequalities by converting to a perfect square which is greater or equal to zero. Factor the difference of two squares the sum of two cubes the difference of two cubes Manipulate exponent notation surds logs Solve equations involving surds by repeated squaring and back check that all roots satisfy the original equation. Solve equations involving logs by exponentiation and back check that all the roots satisfy the original equation. Prove identities with logs using the change of case law. Prove the factor theorem for quadratic and cubic equations. 38

40 Solve problems using the factor theorem by recognizing the connection between factors and roots of the equation. Find the roots of cubic equations where one root is an integer. Solve more difficult problems using the factor theorem when a quadratic factor of a cubic expression is given. Recognize the special relationship between the coefficients and the roots of a quadratic equation and solve problems related to these. In this course a candidate must be comfortable with: Manipulation of Formulae Surds Logs Exponentials Quadratic Equations Simultaneous Equations Solving Inequalities before attempting to move on to move advanced topics. 39

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