Question 2:
2 | (a) | − − 6  ( − 6 ) 2 − 4  1  5 8
x =
2  1
√−196 = 14i
6  1 4 i
x = = 3  7 i
2 | M1
B1FT
A1 | 1.1
1.1
2.5 | DR. Correctly using formula or
completing the square.
Condone “–6 ” for M1.
FT their negative discriminant.
Writing square root of negative
number as the correct multiple
of i.
Must be a + bi. | Condone missing brackets under
root if –6 squares to 36 (ie  = –
196 rather than –268).
“–196” (or “–49”) must be seen.
(a) | Alternative method: | B1 | May be embedded.
Using the fact that the roots of a
The roots are  = a + bi and  = a – bi (where | real quadratic form a complex
a and b are real) | conjugate pair.
 +  = –(–6)/1 = 6 and  = 58/1 = 58
So 2a = 6 (=>a = 3) and a2 + b2 = 58 | M1 | Finding the numerical value of
the sum and product of the roots.
 +  = –(–6)/1 = 6 and  = 58/1 = 58 | Finding the numerical value of | Could also be found by expanding
So 2a = 6 (=>a = 3) and a2 + b2 = 58 | the sum and product of the roots. | (x – (a + bi))(x – (a – bi)) and
comparing with equation.
Could also substitute (a + bi) into
the equation to derive (2ab – 6b) =
0 and a2 – b2 – 6a + 58 = 0.
So b2 = 58 – 9 = 49 so roots are 3  7i | A1 | Must be a + bi. | a real =>b 0 =>a = 3 =>b = 7
[3]
B1
May be embedded.
M1
(b) | ( ) 5 1 2 2  
a r g 1 0 5 1 2 i t a n 1  − + = − =
1 0 3 −
( )
( ) 5 ( )
a r g 1 0 5 1 2 i 5 a r g 1 0 5 1 2 i − + = − + =
2 1 0  
5    =
3 3
2 4  
so required angle is o r   −
3 3 | M1
M1
A1
[3] | 1.1
1.1
1.1 | DR. Using correct formula for
argument of complex number
with non-zero real and
imaginary parts.
Using De Moivre’s Theorem for
their angle. Condone error in
modulus if shown.
cao | 5 1 2 
Condone t a n   =  = − or
1 0 3 −
5 12 2 
tan−1 = or − for M1.
10 3 3
or using a valid method for finding
z5 explicitly (eg by expansion or by
writing 1 0 5 1 2 i 2 0 e 23 i  − + = )
z5 = –1600000 – 16000003 i or
23
( 2 0 5 ) e 5 i  