Mathematics · May / June 2026 · Paper 22 · Question 25

(0.5, 5) and (3, –5) [5]

Allow equivalent fractions for 0.5 e.g. 2 4

M2 for 2x2 – 7x + 3[ = 0] oe three –2x2 + 7x – 3, allow in the form 2x2 – 7x = –3 if next terms step is completing the square.

or M1 for 7 – 4x = 4 + 3x – 2x2 oe their three-term quadratic cannot be 4 + 3x – 2x2 for or better this M1 mark M1FT for correct method to solve Condone partial factorisation, e.g. 2x(x – 3) – [1](x – 3), if followed by correct values their three-term quadratic, e.g. factors, formula, completing the of x. 2x – 1 = 0 oe and x – 3 = 0 oe imply M1. square (2x – 1)(x – 3) [= 0] ( ) ( ) −− ± − 2 −×× ± − ± ± 7 7 4 2 3 oe e.g. 7 49 24 , 7 25 , 7 5 oe × 4 4 4 2 2

B1 for one correct pair of Working must be shown for SC1 coordinates or two correct x-values If eliminating x: or two correct y-values

M2 for y2 – 25 [= 0] oe two terms If B0 scored and at least 2 method − − 2 7 y 7 y = + −

or M1 for y 4 3 2 marks awarded,

SC1 for correct 4 4 substitution of both of their x-

M1 for correct method to solve their quadratic, values into y = 7 – 4x or e.g. factors, formula, completing the square y = 4 + 3x – 2x2 B1 and SC1 equivalent steps to main scheme.