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1 Quadratic Equations Mark Schee 1 Level A LEVEL Ea Board Edecel GCE Subject Matheatics Module Core 1 Topic Algebra and Functions Sub-Topic Quadratic Equations Booklet Mark Schee 1 Tie Allowed: Score: Percentage: 57 inutes / 47 /100 Grade Boundaries: A* A B C D E U >85% 777.5% 70% 6.5% 57.5% 45% <45%

2 Nuber 1(a) b 4ac 0 e.g. 4 4( p 1) ( p 5) 0 or 0 4 4( p 1) ( p 5) or 4 4( p 1) ( p 5) or 4( p 1) ( p 5) 4 4 p 6p 5 p 6p 1 0 p 6p 1 0 p... Schee : Attepts to use b 4ac with at least two of a, b or c correct. May be in the quadratic forula. Could also be, for eaple, coparing or equating b and 4ac. Must be considering the given quadratic equation. Inequality sign not needed for this.there ust be no ters. A1: For a correct un-siplified inequality that is not the given answer Correct solution with no errors that includes an epansion of (p 1)(p 5) For an attept to solve p 6p 1 0 (not their quadratic) leading to solutions for p (do not allow attepts to factorise ust be using the quadratic forula or copleting the square) p 3 or any equivalent correct epressions e.g. p p (May be iplied by their inequalities) Discriinant ust be a single nuber not e.g Allow the A1 to score anywhere for solving the given quadratic : Chooses outside region not dependent on the previous ethod ark A1: p 3 8, p 3 8 or equivalent e.g. p 3 8 or p p, p 3, , Allow,, or or a space between the answers but do not allow p 3 8 and p 3 8 (this scores A0) Apply ISW if necessary. A1 A1* (3) A1 A1 A correct solution to the quadratic followed by p 3 8scores A1M0A0 3 8 p 3 8 scores A0 Allow candidates to use rather than p but ust be in ters of p for the final A1 (4) (7 arks)

3 Nuber Schee.(a) 4 y Allow y or y y or y squared "4 " not required Must be seen in part (a) 8y 9y 1 (8y 1)(y 1) 0 y... or (8( ) 1)(( ) 1) 0... For attepting to solve the given equation as a 3 ter quadratic in y or as a 3 ter quadratic in leading to a value of y or (Apply usual rules for solving the quadratic see general guidance) Allow (or any other letter) instead of y for this ark e.g. an attept to solve Both correct answers of 1 8 (oe) B1 (1) (or y) 1, and 1 for or y or their letter but not unless (or y) is iplied later : A correct attept to find one nuerical value of fro their (or y) which ust have coe fro a 3 ter quadratic equation. If logs are used then they ust be evaluated. A1: Both 3 and/or 0 May be iplied by e.g. 1 3 and 0 1 and no 8 etra values. A1 A1 (4) (5 arks)

4 Nuber 3(a) Schee 4 k(1 ) + 5k( = 0) So + 8k+ k = 0* (8 k) 4k k = 1 (oe) 16 Makes y the subject fro the first equation and substitutes into the second equation (= 0 not needed here) or eliinates y by a correct ethod. Correct copletion to printed answer. There ust be no incorrect stateents. : Use of b 4ac(Could be in the quadratic forula or an inequality, = 0 not needed yet). There ust be soe correct substitution but there ust be no s. No forula quoted followed by e.g. 8k 4k = 0is M0. A1: Correct epression. Do not condone issing brackets unless they are iplied by later work but can be iplied by (8 k) > 4k etc. Cso (Ignore any reference to k = 0) but there ust be no contradictory earlier stateents. A fully correct solution with no errors. A1cso A1 A1 () (3) Way Equal roots + 8k+k = ( + k) 8k = k : Correct strategy for equal roots A1: Correct equation A1 Way 3 k = 1 (oe) Cso (Ignore any reference to k = 0) A1 16 Copletes the Square + 8k+ k = ( + 4 k) 16k +k 16k k = 0 : ( ± 4k) ± p± k, p 0 A1: Correct equation A1 (c) k = 1 (oe) Cso (Ignore any reference to k = 0) A = 0 so ( + ) = 0 = = 1 1 4, y =1 Substitutes their value of k into the given quadratic and attept to solve their or 3 ter quadratic as far as = (ay be iplied by substitution into the quadratic forula) or starts again and substitutes their value of k into the second equation and solves siultaneously to obtain a value for. First A1 one answer correct, second A1 both answers correct Special Case: , 1, 16 = 0 = y = allow A1A0 4 4 A1A1 (3) (3) [8]

5 Nuber 4. (a) Schee Boy s Sequence: 10,15, 0, 5,... { a = 10, d = 5 T 15 = } a + 14d = ( 5); = 80 or (0.05); = 0.80 ; A1 { S = } 60 [ (10) 59(5) ] 60 + A1 = 30(315) = 9450 or A1 Boy s Sister s Sequence: 10, 0, 30, 40,... (c) { a = 10, } ( (10) + ( 1)( 10) ) d = 10 S = or 10( + 1) or 5 ( + 1) A1 63 or 6300 = ( (10) + ( 1)(1 0) ) d 6300 = (10)( +1) or 1600 = 10 ( +1) 160 = ( +1) = ( + 1) (*) A1 cso (d) { = }35 B1 Notes (a) : for using the forula a + 14d with either a or d correct. A1: for 80 or 80p or 0.80 or 0.80p and apply ISW. Otherwise, 80 or 0.80 or 0.80p would be A0. Award M0 if candidate applies a + 59d. Listing the first 15 ters and highlighting that the 15 th ter is 80 or listing 15 ters with the final 15 th ter aligned with 80 will then be awarded all two arks of A1. Writing down 80 with no working is A1. [] [3] [4] [1] 10 : for use of correct 60 [ (10) + 59(5) ] or 15 ( (10) + 14(5) ) with a = 10, d =5 and n = 60 or a = 10, d = 5 and n = 15. If a candidate uses n ( a + l) with n = 60 or 15, there ust be a full ethod of finding or stating l as either a + 59d ( =305) or a + 14d( = 80 ), respectively. [ ] or 60 [ (0.1) ] 1 st A1: for a correct epression for S 60. ie. 60 (10) + 59(5) + 59(0.05) or [ ] or [ ]. This ark can be iplied by later working. nd A1: for 9450 or 9450p or and apply ISW. Otherwise, 9450 or without sign is A0. Note: the bracketing error of 60 (10) + 59(5) is A0 unless recovered fro later working. Adding together the first 60 ters to obtain 9450 will then be awarded all three arks of A1A1.

6 (c) 1 st : for correct use of S forula with one of a or d correct. 1 st A1: for a correct epression for S. Eg: ( (10) + ( 1)(10) ) or nd : for foring a suitable equation using 63 or 6300 and their S. 10( + 1) or 5 ( + 1) Dependent on 1st. nd A1cso: for reaching the printed result with no incorrect working seen. Long ultiplication is not necessary for the final accuracy ark. Note: ( (10) + ( 1)(10) ) = 630 and not either 6300 or 63 is dm0. Beware: Soe candidates will try and fudge the result given on the question paper. Notes for awarding nd A1 Going fro ( + 1) = 160 straight to + ( 1) = is nd A1. Going fro ( + 1) = soe factor decoposition of 6300 straight to ( + 1) = is nd A1. Going fro 10 ( + 1) =1600 straight to + ( 1) = is nd A0. Going fro ( + 1) = straight to + ( 1) = is nd A0. Alternative: working in an different letter, say n or p. A1: for n n ( (10) + (n 1)(10) ) (although iing letters eg. ( (10) + ( 1)(10) ) is M0A0). d: for 63 or 6300 = n ( (10) + (n 1)(1 0) ) n Leading to 6300 = (10)( n+ 1) 160 = n ( n+ 1) = n( n+ 1) The candidate then needs to write either = ( +1) or n or = n to gain the final A1. (d) B1: for 35 only.

7 Nuber Schee 5. (a) Discriinant: b 4ac = (k + 3) 4k or equivalent A1 (k + 3) 4k = k + k + 9 = (k +1) + 8 A1 (c) For real roots, b 4ac 0 or b 4ac > 0 or ( k + 1) + 8 > 0 (k +1) 0 for all k, so b 4ac > 0, so roots are real for all k (or equiv.) Notes (a) : attept to find discriinant substitution is required If forula b 4ac is seen at least of a, b and c ust be correct If forula b 4ac is not seen all 3 of a, b and c ust be correct Use of b + 4ac is M0 A1: correct unsiplified : A pt at copletion of square (see earlier notes) A1: both correct (no ft for this ark) (c) : States condition as on sche or attepts to eplain that their ( k + 1) + 8 is greater than 0 A1: The final ark (A1cso) requires (k +1) 0 and conclusion. We will allow ( k + 1) > 0 ( or word positive) also allow b A1 cso () () () 6 4ac 0and conclusion.

8 Schee nuber 6. (a) [No real roots iplies b 4ac< 0.] b 4ac= q 4 q ( 1 ) So q 4 q ( 1)<0 i.e. q + 8 q <0 (*) A1cso () ( ) q(q + 8) = 0 or q ± 4 ± 16=0 (q) = 0 or 8 ( cvs) A1 8< q < 0 or q (-8, 0) or q < 0 and q > -8 A1ft (3) 5 (a) for attepting b 4ac with one of b or a correct. < 0 not needed for This ay be inside a square root. A1cso for siplifying to printed result with no incorrect working or stateents seen. Need an interediate step e.g. q 8q<0 or q 4 q 1<0 or q 4(q)( 1) <0 or i. e. ust have or brackets on the 4ac ter < 0 ust be seen at least one line before the final answer. q 8q( 1) <0 or q 8q 1<0 for factorizing or copleting the square or attepting to solve q ± 8q= 0. A ethod that would lead to values for q. The = 0 ay be iplied by values appearing later. st 1 A1 for q = 0 and q = 8 nd A1 for 8<q < 0. Can follow through their cvs but ust choose inside region. q < 0, q > 8 is A0, q < 0 or q > 8 is A0, (-8, 0) on its own is A0 BUT q < 0 and q > 8 is A1 Do not accept a nuber line for final ark

9 Schee nuber 7. (a) Attept to use discriinant b 4ac k 4(k + 3) > 0 k 4k 1 > 0 (*) A1cso () k 4k 1 = 0 ( k± a) k± ( b), with ab 1 or ( k =) k = and 6 k <, k > 6 or (, );(6, ) 4± = or ( k ) ± 1 (both) A1 M: choosing outside A1ft (4) (a) for use of b 4ac, one of b or c ust be correct. Or full attept using copleting the square that leads to a 3TQ in k e. k k + = (k+3) 4 A1cso Correct arguent to printed result. Need to state (or iply) that b 4ac >0 and no incorrect working seen. Must have >0. If > 0 just appears with k 4( k+ 3)>0that is OK. If >0 appears on last line only with no eplanation give A0. b 4ac followed by k 4k 1 > 0 only is insufficient so M0A0 6 e.g. k 4 1 k+3 (issing brackets) can get A0 but Using b 4 ac > 0 is M0. k + 4( k+ 3) is M0A0 (wrong forula) st st 1 A1 nd nd A1f.t. for attepting to find critical regions. Factors, forula or copleting the square. for k = 6 and only for choosing the outside regions as printed or f.t. their (non identical) critical values 6 < k < is A0 but ignore if it follows a correct version < k < 6 is M0A0 whatever their diagra looks like Condone use of instead of k for critical values and final answers in. Treat this question as 3 two ark parts. If part (a) is seen in or vice versa arks can be awarded.