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bg-background","children":["$","$L3b",null,{"backHref":"/igcse/maths/oxford-aqa/questions","basePath":"/igcse/maths/oxford-aqa/grade-8-9-8-1-quadratic-simultaneous-equations/questions","formulaUrl":"https://www.oxfordaqa.com/oaqaresources/maths/assessment/9260-oxfordaqa-international-gcse-mathematics-formulae-sheet-ext-e3.pdf","initialIndex":3,"initialQuestionWindow":[{"id":"7a998f38-9f5d-49d6-bf36-a741cb70359a","revisionId":"456595a5-53e6-410a-91ce-7754283d6178","questionNumber":"4","questionType":"LA","specification":"","difficulty":2,"difficultyBand":"hard","totalMarks":5,"sourceTitle":"Quadratic Simultaneous Equations","calculatorUse":null,"isCalculation":true,"parts":[{"id":"fa046038-46e5-4d72-80d3-1284097d3cd5","label":"0","content":"Solve the simultaneous equations\n\n$$x^2 + y^2 = 20$$\n\n$$3x = 2 - y$$","markscheme":"- Rearranges linear to $y = 2 - 3x$ and substitutes: $x^2 + (2 - 3x)^2 = 20$ M1\n- Simplifies to $10x^2 - 12x - 16 = 0$ or $5x^2 - 6x - 8 = 0$ M1\n- Factorises to $(5x + 4)(x - 2) = 0$ M1\n- Finds $x = -0.8$ and $x = 2$ A1\n- Finds $y = 4.4$ and $y = -4$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Rearranges linear equation and substitutes into the quadratic.","code":"M"},{"id":"p2","marks":1,"description":"Simplifies to a quadratic equation equal to zero.","code":"M"},{"id":"p3","marks":1,"description":"Solves the quadratic equation (factorisation or formula).","code":"M"},{"id":"p4","marks":1,"description":"Finds both correct values for $x$.","code":"A"},{"id":"p5","marks":1,"description":"Finds both correct values for $y$.","code":"A"}],"finalAnswers":["x = -0.8, y = 4.4; x = 2, y = -4"],"generatedAt":"2026-05-07T02:12:13.942Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitute_for_y","label":"Substitute for y","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3x=2-y$"},{"latex":"$$-y=3x-2$"},{"latex":"$$y=2-3x$"}]},"order":0,"title":"Make y the subject","description":"Rearrange the linear equation so that it can be substituted into the equation containing the squares. From $-y=3x-2$, multiply both sides by $-1$ to obtain $y=2-3x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+y^2=20$"},{"latex":"$$x^2+(2-3x)^2=20$"}]},"order":1,"title":"Substitute for y","description":"Replace $y$ in the circle equation with $2-3x$. Writing this substitution earns the first method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+(2-3x)^2=20$"},{"latex":"$$x^2+4-12x+9x^2=20$"},{"latex":"$$\\color{@amber}{\\htmlData{anchor=quadratic_full}{10x^2-12x-16=0}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=quadratic_simple}{5x^2-6x-8=0}}$"}],"connections":[{"to":"quadratic_simple","from":"quadratic_full","color":"sky","label":"divide throughout by 2"}]},"order":2,"title":"Form the quadratic","description":"Expand the square carefully, collect terms, then divide throughout by $2$. Reaching either quadratic shown earns the second method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x^2-6x-8=0$"},{"latex":"$$5x^2-10x+4x-8=0$"},{"latex":"$$5x(x-2)+4(x-2)=0$"},{"latex":"$$(5x+4)(x-2)=0$"}]},"order":3,"title":"Factorise the quadratic","description":"Split the middle term, then factorise by grouping. Reaching $(5x+4)(x-2)=0$ earns the third method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x+4=0$"},{"latex":"$$x=-\\frac45=-0.8$"},{"latex":"$$x-2=0$"},{"latex":"$$x=2$"}]},"order":4,"title":"Solve both factors","description":"Set each factor equal to zero. Obtaining both roots, $x=-0.8$ and $x=2$, earns the accuracy mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=-0.8:\\quad y=2-3(-0.8)=4.4$"},{"latex":"$$(x,y)=(-0.8,4.4)$"},{"latex":"$$x=2:\\quad y=2-3(2)=-4$"},{"latex":"$$(x,y)=(2,-4)$"}]},"order":5,"title":"Find both y-values","description":"Use $y=2-3x$ once for each root. Repeat for the other root, keeping each $x$ paired with its corresponding $y$; this earns the final accuracy mark."}]}],"generatedAt":"2026-07-20T16:46:49.804Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":{"percent":58},"walkthroughVideo":null},{"id":"0cc49e8e-d89b-426d-ad24-b2f180fd7bba","revisionId":"3445ce4f-71ae-4a2a-a31c-aac486e40c91","questionNumber":"6","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","totalMarks":5,"sourceTitle":"Quadratic Simultaneous Equations","calculatorUse":null,"isCalculation":true,"parts":[{"id":"6b5c4dff-11d3-4f86-adff-5a3c886c704f","label":"0","content":"Solve the simultaneous equations \nGive your answers to 3 significant figures\n\n$$x^2 + y^2 = 20$$\n\n$$2x + y = 3$$","markscheme":"- Rearranges linear to $y = 3 - 2x$ and substitutes: $x^2 + (3 - 2x)^2 = 20$ M1\n- Simplifies to $5x^2 - 12x - 11 = 0$ M1\n- Uses quadratic formula: $x = \\frac{12 \\pm \\sqrt{(-12)^2 - 4(5)(-11)}}{10}$ M1\n- Finds $x = 3.11$ and $x = -0.708$ A1\n- Finds $y = -3.22$ and $y = 4.42$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Substitutes linear into quadratic and expands brackets.","code":"M"},{"id":"p2","marks":1,"description":"Simplifies to a quadratic equation $5x^2 - 12x - 11 = 0$.","code":"M"},{"id":"p3","marks":1,"description":"Applies the quadratic formula correctly.","code":"M"},{"id":"p4","marks":1,"description":"Finds both $x$ values to 3 significant figures.","code":"A"},{"id":"p5","marks":1,"description":"Finds both $y$ values to 3 significant figures.","code":"A"}],"finalAnswers":["x = 3.11, y = -3.22; x = -0.708, y = 4.42"],"generatedAt":"2026-05-07T02:12:13.942Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution_quadratic_formula","label":"Substitution and quadratic formula","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@amber}{\\htmlData{anchor=linear}{2x+y=3}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=subject}{y=3-2x}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"linear"},{"color":"sky","style":"text-color","anchor":"subject"}],"connections":[{"to":"subject","from":"linear","color":"sky","label":"subtract 2x"}]},"order":0,"title":"Make y the subject","description":"Rearrange the linear equation so that it can be substituted into the circle equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+\\htmlData{anchor=original_y}{\\color{@orange}{y^2}}=20$"},{"latex":"$$x^2+\\htmlData{anchor=substituted_bracket}{\\color{@sky}{(3-2x)^2}}=20$"}]},"order":1,"title":"Substitute for y","description":"Replace y in the first equation with the whole expression 3 − 2x. Rearranging to y = 3 − 2x and then substituting to obtain x² + (3 − 2x)² = 20 together earn the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=source_equation}{x^2+(3-2x)^2=20}$"},{"latex":"$$\\htmlData{anchor=expanded_equation}{x^2+9-12x+4x^2=20}$"},{"latex":"$$\\htmlData{anchor=quadratic_equation}{5x^2-12x-11=0}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"source_equation"},{"color":"sky","style":"text-color","anchor":"expanded_equation"},{"color":"green","style":"text-color","anchor":"quadratic_equation"}]},"order":2,"title":"Form the quadratic","description":"First expand (3 − 2x)² to obtain x² + 9 − 12x + 4x² = 20. Then collect the x² terms and move everything to one side to obtain 5x² − 12x − 11 = 0. Obtaining this quadratic earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x=\\dfrac{-b\\pm\\sqrt{b^2-4ac}}{2a}$"},{"latex":"$$x=\\color{@sky}{\\htmlData{anchor=substitution}{\\dfrac{12\\pm\\sqrt{(-12)^2-4(5)(-11)}}{10}}}$"},{"latex":"$$x=\\color{@green}{\\htmlData{anchor=discriminant}{\\dfrac{12\\pm\\sqrt{364}}{10}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"substitution"},{"color":"green","style":"text-color","anchor":"discriminant"}],"connections":[{"to":"discriminant","from":"substitution","color":"sky","label":"evaluate the discriminant"}]},"order":3,"title":"Use the formula","description":"For 5x² − 12x − 11 = 0, use a = 5, b = −12 and c = −11. This correct formula setup earns the third M1."},{"type":"text","order":4,"title":"Round both x-values","content":"Evaluating both signs gives $x=3.107878\\ldots$ or $x=-0.707878\\ldots$. To **3 significant figures**, these are\n\n$$x=3.11 \\quad \\text{or} \\quad x=-0.708.$$\n\nBoth values are needed because the quadratic has two solutions. These correctly rounded $x$-values earn A1.","highlights":[{"text":"Give your answers to 3 significant figures","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@orange}{\\htmlData{anchor=xpositive}{x=3.107878\\ldots}}$"},{"latex":"$$\\color{@sky}{\\htmlData{anchor=ynegative}{y=3-2(3.107878\\ldots)=-3.215756\\ldots}}$"},{"latex":"$$\\color{@purple}{\\htmlData{anchor=xnegative}{x=-0.707878\\ldots}}$"},{"latex":"$$\\color{@green}{\\htmlData{anchor=ypositive}{y=3-2(-0.707878\\ldots)=4.415756\\ldots}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"xpositive"},{"color":"sky","style":"text-color","anchor":"ynegative"},{"color":"purple","style":"text-color","anchor":"xnegative"},{"color":"green","style":"text-color","anchor":"ypositive"}],"connections":[{"to":"ynegative","from":"xpositive","color":"sky","label":"substitute into y = 3 − 2x"},{"to":"ypositive","from":"xnegative","color":"green","label":"substitute into y = 3 − 2x"}]},"order":5,"title":"Find the matching y-values","description":"Substitute each unrounded x-value into y = 3 − 2x. Keep the pairs matched; rounding x too early can reduce accuracy."},{"type":"text","order":6,"title":"State the solution pairs","content":"Rounding each $y$-value to $3$ significant figures gives the two solutions\n\n$$\\boxed{(x,y)=(3.11,-3.22)}$$\n\nand\n\n$$\\boxed{(x,y)=(-0.708,4.42)}.$$\n\nWriting the matching $y$-values correctly earns the final A1."}]}],"generatedAt":"2026-07-20T16:53:22.734Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":{"percent":43},"walkthroughVideo":null}],"subjectCode":"igcse_maths__oxford_aqa","summaries":[{"id":"121b4e20-35c5-4163-bf65-e301615e6d2d","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"d8b1c68f-6bd3-4b56-9555-439c1a785cde","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"7d960de9-c383-4ca6-b926-a8586de290be","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"7a998f38-9f5d-49d6-bf36-a741cb70359a","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"57f2dd00-7fa2-491a-b4ff-5a0332e20c8e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"0cc49e8e-d89b-426d-ad24-b2f180fd7bba","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"5a91a9ec-0d6f-4980-aa55-d2064da49327","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"05721abb-3610-49f7-aaa6-e71c8d3116a4","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a16bde18-5d62-42e0-a851-f3075c4dd2f6","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true}],"title":"Quadratic Simultaneous Equations","topicId":"f3d8a89f-d75f-41ac-8628-e593ef6f374e","topicSidebar":"$L3c","topicSlug":"grade-8-9-8-1-quadratic-simultaneous-equations","topicTitle":"Quadratic Simultaneous Equations"}]}],"$L3d","$L3e"]}]

Quadratic Simultaneous Equations

Question 1

Quadratic Simultaneous Equations Questions

Quadratic Simultaneous Equations

Question 1

Quadratic Simultaneous Equations Questions