In the context of basic factorisation, solve the inequality x2 - 5x + 6 > 0 for x on the real number line.
(Total for Question 1 is 1 mark)
2
For the parabola context y = x2 - x - 6, find where y ≤ 0. Give your answer in inequality form for x.
(Total for Question 2 is 1 mark)
3
On the real number line, solve 2x2 - 8x < 0 by factorising and state the solution set.
(Total for Question 3 is 2 marks)
4
Consider the quadratic expression q(x) = -x2 + 6x - 5. Solve q(x) ≥ 0 and give the solution for x.
(Total for Question 4 is 2 marks)
5
Sketch and shade the region on the xy-plane for the inequality y > x2 - 4x + 3. Indicate clearly which side of the parabola is shaded and mark the critical points on the x-axis.
(Total for Question 5 is 3 marks)
6
A quadratic region problem: solve x2 + x - 12 ≤ 0 and then sketch the x-axis interval to show the solution set clearly on a number line.
(Total for Question 6 is 3 marks)
7
Solve the inequality 3x2 - 18x + 23 > 0. The roots are not integers. Show your working including the critical values found using the quadratic formula, and give the final solution in inequality notation.
(Total for Question 7 is 4 marks)
Mark scheme · IG.M18 Quadratic Inequalities: Solving and Graphing Regions
Question 1
B1 correct solution x < 2 or x > 3, in correct inequality notation
Answer: x < 2 or x > 3
Question 2
B1 -2 ≤ x ≤ 3 cao
Answer: -2 ≤ x ≤ 3
Question 3
M1 factorises to 2x(x - 4) and identifies roots x = 0 and x = 4
A1 0 < x < 4, correct inequality notation
Answer: 0 < x < 4
Question 4
M1 rearranges or factorises, noting -x2 + 6x - 5 = -(x2 - 6x + 5) and finds roots x = 1 and x = 5
A1 1 ≤ x ≤ 5 cao
Answer: 1 ≤ x ≤ 5
Question 5
M1 identifies and plots the roots of x2 - 4x + 3 = 0 as x = 1 and x = 3 or marks points (1,0) and (3,0)
M1 sketches a correct upward-opening parabola passing through the roots
A1 shades the region above the parabola (y > x2 - 4x + 3) clearly
Answer: Region above the parabola y = x2 - 4x + 3, excluding the curve itself, with the curve crossing the x-axis at x = 1 and x = 3
Question 6
M1 factorises to (x + 4)(x - 3) and finds roots x = -4 and x = 3
A1 -4 ≤ x ≤ 3, correct inequality notation
B1 number line or small sketch showing closed dots at -4 and 3 and shading between them
Answer: -4 ≤ x ≤ 3
Question 7
M1 uses quadratic formula correctly and computes discriminant: b2 - 4ac = 324 - 276 = 48
M1 finds critical values x = (18 ± √48)/6 and simplifies exactly to x = (9 ± 2sqrt(3))/3, oe
A1 gives final solution in correct form: x < (9 - 2sqrt(3))/3 or x > (9 + 2sqrt(3))/3, cao
A1 states approximate critical values, for example x approximately 1.845 and x approximately 4.155, supporting the correct inequality notation
Answer: x < (9 - 2sqrt(3))/3 or x > (9 + 2sqrt(3))/3