":"Question Set 3: Equations and Inequalities","calculatorUse":null,"parts":[{"id":"7ce371fd-6fc5-4a92-a54d-c42d931ecc83","label":"0","content":"Using algebra, solve the inequality $x^2 - x - 12 < 0$ writing your answer in set notation.","markscheme":"$$(x - 4)(x + 3) < 0$\nCritical values: $x = 4, x = -3$\n![](https://assets.mathsgenie.co.uk/booklet/fb60f375c562fab46b8fe630218c96fc60a2d7222d64157fd37948813997371c.jpg)\n$\\{x: -3 < x < 4\\}$ 3","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Factorises quadratic to $(x - 4)(x + 3)$.","code":"M"},{"id":"p2","marks":1,"description":"Identifies critical values $-3$ and $4$.","code":"A"},{"id":"p3","marks":1,"description":"Correct answer in set notation $\\{x: -3 < x < 4\\}$.","code":"A"}],"finalAnswers":["\\{x: -3 < x < 4\\}"],"generatedAt":"2026-05-12T02:14:06.552Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"factor_sign_chart","label":"Factorise and use signs","steps":[{"type":"text","order":0,"title":"Factorise the quadratic","content":"Factorise the quadratic expression:\n\n$$x^2-x-12=(x-4)(x+3).$$\n\nThis earns the **first mark**. The inequality is now a product, so we can find where its sign changes.","highlights":[{"text":"Using algebra, solve the inequality $x^2 - x - 12 < 0$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=factorised}{\\color{@sky}{(x-4)(x+3)<0}}"},{"latex":"\\htmlData{anchor=firstfactor}{\\color{@amber}{x-4=0}}"},{"latex":"\\htmlData{anchor=secondfactor}{\\color{@amber}{x+3=0}}"},{"latex":"\\htmlData{anchor=firstroot}{\\color{@green}{x=4}}"},{"latex":"\\htmlData{anchor=secondroot}{\\color{@green}{x=-3}}"}]},"order":1,"title":"Find the critical values","description":"Set each factor equal to zero. Finding the critical values x = 4 and x = -3 earns the second mark."},{"type":"text","order":2,"title":"Check every interval","content":"The critical values split the number line into three intervals.\n\n- For $x<-3$, both factors are negative, so their product is positive.\n- For $-34$, both factors are positive, so their product is positive.\n\nTherefore, only the middle interval satisfies $(x-4)(x+3)<0$.","highlights":[{"text":"$$x^2 - x - 12 < 0$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=interval}{\\color{@sky}{-3M1 A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Factors out $3x$ to get $3x(25 - 4x^2)$.","code":"M"},{"id":"p2","marks":1,"description":"Uses difference of two squares to factorise completely.","code":"A"}],"finalAnswers":["3x(5+2x)(5-2x)"],"generatedAt":"2026-05-12T02:14:06.555Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"factorising","label":"Factorise completely","steps":[{"type":"hyper_la

1.1 Algebra and functions

Question 12

1.1 Algebra and functions Questions

1.1 Algebra and functions

Question 12

1.1 Algebra and functions Questions