tions","basePath":"/gcse/maths/edexcel/grade-8-9-8-6-quadratic-inequalities/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/gcse-maths/gcse-formulae-to-remember.pdf","initialIndex":11,"initialQuestionWindow":[{"id":"b77033cf-d912-4f57-92b2-f8dcb3b4cc5c","revisionId":"82bd4867-b175-4877-9514-7067ea0ffa43","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"9f4a0595-b12a-47f2-9d05-5e5f6c534493","label":"","content":"Solve $y^2 < 64$","markscheme":"- Rearranges to $y^2 - 64 < 0$ M1\n- Factorises using difference of two squares: $(y+8)(y-8) < 0$ M1\n- Identifies critical values: $y = -8$ and $y = 8$ M1\n- States correct inequality: $-8 < y < 8$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Rearranges the inequality to the standard quadratic form $y^2 - 64 < 0$.","code":"M"},{"id":"p2","marks":1,"description":"Factorises the quadratic expression using the difference of two squares.","code":"M"},{"id":"p3","marks":1,"description":"Identifies the critical values as $y = -8$ and $y = 8$.","code":"M"},{"id":"p4","marks":1,"description":"States the final solution as a single inequality range $-8 < y < 8$.","code":"A"}],"finalAnswers":["-8 < y < 8"],"generatedAt":"2026-08-10T10:52:35.243Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"bf84f01a-7ff1-4b76-9040-432f9771b4dd","revisionId":"2441d34c-cd6f-420c-be4e-e9db004ea950","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"fd42413e-d92f-40c2-94a1-bab1d82221e7","label":"a","content":"Solve $x^2 - 2x - 15 \\leq 0$","markscheme":"- Factorises the quadratic: $(x - 5)(x + 3) \\leq 0$ M1\n- Identifies critical values: $x = 5$ and $x = -3$ M1\n- States correct inequality: $-3 \\leq x \\leq 5$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Factorises the quadratic expression correctly or shows use of quadratic formula with correct values.","code":"M"},{"id":"p2","marks":1,"description":"Identifies the correct critical values from the factorised expression.","code":"M"},{"id":"p3","marks":1,"description":"States the final solution in the correct inequality form.","code":"A"}],"finalAnswers":["-3 \\leq x \\leq 5"],"generatedAt":"2026-08-10T10:54:14.480Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"c653f59d-5e40-499e-b72f-28eafdf0ed91","revisionId":"ea4fe8ff-f67e-456f-a6aa-3eb6cf20894d","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"eed9fd4d-5346-4794-894c-925b18ba7744","label":"a","content":"Solve $x^2 < 2x + 15$","markscheme":"- Rearranges to $x^2 - 2x - 15 < 0$ M1\n- Factorises the quadratic: $(x - 5)(x + 3) < 0$ M1\n- Identifies critical values: $x = 5$ and $x = -3$ M1\n- States correct inequality: $-3 < x < 5$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Rearranges to $x^2 - 2x - 15 < 0$","code":"M"},{"id":"p2","marks":1,"description":"Factorises the quadratic correctly into two brackets","code":"M"},{"id":"p3","marks":1,"description":"Identifies critical values: $x = 5$ and $x = -3$","code":"M"},{"id":"p4","marks":1,"description":"States correct final inequality solution","code":"A"}],"finalAnswers":["-3 < x < 5"],"generatedAt":"2026-08-10T10:52:49.265Z","modelVersion":"gemini-3-flash-preview"},"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null}],"

Quadratic Inequalities Questions

Quadratic Inequalities Questions

Quadratic Inequalities Questions

Quadratic Inequalities Questions