1 b Uses α + β = to write 4p = 6 a TBC Solves to find 3 p = Uses c αβ = to write a 30 3p = k Solves to find k = 40 9 (4) (4 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1
a b m Uses α + β = to write Re( α ) = a M1.a TBC Solves to find m = 40 () b Uses c a αβ = to write Re( α) ( ) ( Im ( α) ) + = n M1.a Solves to find n = 80 and concludes n > 80 () (4 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.
3a k States αβ + βγ + γα =, αβγ = 10 and α + β + γ = 11 B1 1.1b TBC (1) b Deduces that α * = 1 7i is a root. M1.a αα * = 1+ 7 1 7 = 0 Finds ( i)( i) Uses αβγ = 10 αα * γ = 10 to state 1 γ = c k Uses αβ + βγ + γα = to write 1 ( ) 1 k 0 + 1 7i + ( 1+ 7i) = (3) M1.a Solves to find k = () (6 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3
4a Multiplies the three given roots together and sets the result equal 1 13 to or. For example ( α ) α + + 46 = or α α 1 13 ( α ) α + + 46 = is seen. α α Correctly uses ( α ) α 6α + 13 = 0 1 13 α + + 46 = to find α α Attempts to solve this quadratic using either completing the square or the quadratic formula. TBC Correctly finds α = 3± i States that the roots of f( ) 0 z = are 3+, i 3, i 4 A1.a () b Applies the process of finding in part (a)) to attempt to find m. (of their three roots found Correctly finds m = 10 Applies the process of using the pair sums to find the value of n. For example, 13 + 6γ = n is seen. Correctly finds n = 37 (4) (9 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4
a States or implies that γ = 4i is a root. M1.a TBC 4 Uses α + β + γ + δ = α + β + 4 = to write 4 α + β = 0 αβγ + αβδ + αγδ + βγδ Uses 76 = αβ ( + 4i) + αβ ( 4i) + α ( 0) + β ( 0) = 4 4αβ + 0 α + β = 69 to write ( ) A1 3.1a A1 3.1a b Makes an attempt to solve for α and β, for example α = β is substituted into 4αβ + 0( α + β ) = 69 Forms a quadratic in α or β : 4α 8α 9 0 + = or 4β 8β + 9 = 0 or equivalent is seen and attempts to solve the quadratic. States either α = 1+ i or β = 1+ i (3) M1.a C States the roots of the equation are: + 4, i 4, i 1 + i, 1 i Makes an attempt to use n αβγδ = to find n a A1.a (4) Finds n = 80 () (9 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.
6a bi bii 6 10 States α + β + γ = =, αβ + βγ + γα = and 3 3 0 0 αβγ = = 3 3 α β γ αβγ 4 4 4 Makes an attempt to use ( ) 4 4 4 4 4 4 0 160 000 Finds α β γ ( αβγ ) Makes an attempt to use = = = = 3 81 ( ) ( ) α + β + γ = α + β + γ αβ + βγ + γα ( ) ( ) α + β + γ = α + β + γ αβ + βγ + γα Finds 10 3 = ( ) = 3 3 B1 1.1b TBC (1) () () biii Makes an attempt to multiply out ( α)( β)( γ) Finds or states ( α)( β)( γ) 8 4( ) ( ) = α + β + γ + αβ + βγ + γα αβγ Finds 10 0 40 ( α)( β)( γ) = 8 4( ) + = 3 3 3 (3) (9 marks) Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6
7 States w= x 1 x= w + 1 B1 3.1a TBC States 3 w+ 1 w+ 1 w+ 1 4 + 6 9= 0 Makes an attempt to manipulate the equation into the form 3 aw + bw + cw + d = 0 At least two of a, b, c or d are correct. Fully correct final equation: 3 w w w + 7 7 = 0 () ( marks) 7: Accept an equation that is a multiple of 3 w w w + 7 7 = 0, most likely See also alternative method for first three marks on next page. 3 w w 14w 4 0 + =. Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7
ALTERNATIVE METHOD FOR FIRST THREE MARKS 7 States α + β + γ =, αβ βγ γα 3 + + = and ( α 1 ) + ( β 1 ) + ( γ 1 ) Sum of roots: ( α β γ) ( ) = + + 3= 3= 1 9 αβγ = Pair sum: α 1 β 1 + β 1 γ 1 + γ 1 α 1 ( )( ) ( )( ) ( )( ) = 4( αβ + βγ + γα ) 4( α + β + γ ) + 3 = 43 ( ) 4 ( ) + 3= 7 Product: Applies: ( α 1)( β 1)( γ 1) ( αβγ ) ( αβ βγ γα ) ( α β γ ) = 8 4 + + + + + 1 9 = 8 4( 3) + ( ) 1 = 7 w 3 (their sum roots)w + (their pair sum)w their product = 0 B1 3.1a Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8
8 w States w= x x= B1 3.1a TBC States 4 w w w 6 + 16 1 = 0 Makes an attempt to manipulate the equation into the form 4 3 pw + qw + rw + sw + t = 0. At least two of p, q, r, s and t are correct. Fully correct final equation: 4 w w w 1 + 64 8 = 0 () ( marks) See also alternative method for first three marks on next page. Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9
ALTERNATIVE METHOD FOR FIRST THREE MARKS 8 States α + β + γ + δ = 0, αβ + βγ + γα + γδ + αδ + βδ = 3, 1 αβγ + αβδ + αγδ + βγδ = 8 and αβγδ = Sum of roots: α + β + γ + δ = 0 ( α)( β) + ( β)( γ) + ( γ)( α) + ( )( ) ( )( ) ( )( ) ( ) = 4( 3) = 1...... + γ δ + α δ + β δ Pair sum: = 4 αβ + βγ + γα + γδ + αδ + βδ ( α)( β)( γ) + ( α)( β)( δ) + ( )( )( ) ( )( )( ) ( ) = 8( 8) = 64...... + α γ δ + β γ δ Triple sum: = 8 αβγ + αβδ + αγδ + βγδ 1 = 16 = 16 = 8 Product: ( α )( β )( γ ) ( δ ) ( αβγδ ) Applies: w 4 (their sum roots)w 3 + (their pair sum)w (their triple sum)w + their product = 0 B1 3.1a TBC Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10