mial, set degree to 2. Enter a=1, b=3, c=-25 and solve."},"ti84":{"instructions":"Press [APPS], select 'PlySmlt2', then 'Polynomial Root Finder'. Set Order to 2, press [NEXT]. Enter a=1, b=3, c=-25, and press [SOLVE]."},"desmos":{"expression":"y = x^2 + 3x - 25","instructions":"Type the equation into the input bar to find the x-intercepts of the graph."},"description":"Find the positive root of the quadratic equation using a calculator."}]}]},{"id":"119f33b7-d44e-4270-ba54-efe045c16900:subtraction_method","label":"(b) Subtraction Method","steps":[{"type":"text","order":0,"title":"Set up the subtraction","content":"To find the area of the compound shape using the **Subtraction Method**, we imagine a large outer bounding rectangle and subtract the missing top-right corner rectangle.\n\nFirst, let's find the total width and total height of the bounding box:\n* **Total Width:** $(x - 1) + (2x - 5) = 3x - 6$\n* **Total Height:** $3x + 2$\n\nNext, the missing top-right corner rectangle has:\n* **Width:** $2x - 5$\n* **Height:** $(3x + 2) - 5 = 3x - 3$\n\nNow, we can express the total area of the shape as:\n$$\\text{Area} = \\text{Area of Bounding Rectangle} - \\text{Area of Missing Rectangle}$$"},{"type":"text","order":1,"title":"Expand and simplify","content":"Let's expand both area expressions:\n\n1. **Bounding Rectangle Area:**\n $$(3x - 6)(3x + 2) = 9x^2 + 6x - 18x - 12 = 9x^2 - 12x - 12$$\n\n2. **Missing Rectangle Area:**\n $$(2x - 5)(3x - 3) = 6x^2 - 6x - 15x + 15 = 6x^2 - 21x + 15$$\n\nSubtract the missing rectangle from the bounding rectangle:\n$$\\text{Area} = (9x^2 - 12x - 12) - (6x^2 - 21x + 15)$$\n$$\\text{Area} = 3x^2 + 9x - 27$$"},{"type":"text","order":2,"title":"Form the quadratic equation","content":"Since we are given that the area of the shape is $48\\text{ cm}^2$, we set our simplified expression equal to $48$:\n$$3x^2 + 9x - 27 = 48$$\n\nSubtract $48$ from both sides to form a quadratic equation equal to zero:\n$$3x^2 + 9x - 75 = 0$$\n\nWe can simplify this by dividing the entire equation by $3$:\n$$x^2 + 3x - 25 = 0$$"},{"type":"text","order":3,"title":"Apply the quadratic formula","content":"To solve $x^2 + 3x - 25 = 0$, we use the quadratic formula where $a = 1$, $b = 3$, and $c = -25$ M1:\n\n$$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$\n\nSubstituting our values in:\n$$x = \\frac{-(3) \\pm \\sqrt{(3)^2 - 4(1)(-25)}}{2(1)}$$\n$$x = \\frac{-3 \\pm \\sqrt{9 + 100}}{2}$$\n$$x = \\frac{-3 \\pm \\sqrt{109}}{2}$$"},{"type":"text","order":4,"title":"Solve and select positive solution","content":"Calculating the two possible values of $x$ A1:\n* $x = \\frac{-3 + \\sqrt{109}}{2} \\approx 3.72\\text{ cm}$ (to 3 s.f.)\n* $x = \\frac{-3 - \\sqrt{109}}{2} \\approx -6.72\\text{ cm}$ (to 3 s.f.)\n\nSince $x$ represents a physical measurement, the lengths of the sides must be positive (for example, the width of the left rectangle is $x - 1$, which must be greater than $0$). Therefore, we reject the negative solution B1:\n\n$$x = 3.72$$"}]}],"generatedAt":"2026-06-12T12:14:01.711Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"379da0c0-cf3f-4641-99d2-7877c83d3072","revisionId":"eff35be2-f475-42bd-a039-8f49040bf47b","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":"calc","parts":[{"id":"f77ed271-a829-46d7-85b9-b0173a708d51","label":"a","content":"Solve $3x^2 + 10x + 2 = 0$\n\nGive your solutions correct to 2 decimal places.","markscheme":"- Identifies $a = 3, b = 10, c = 2$ B1\n- Substitutes into the quadratic formula: $x = \\frac{-(10) \\pm \\sqrt{(10)^2 - 4(3)(2)}}{2(3)}$ M1\n- Gives solutions as $-0.21$ and $-3.12$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Identifies $a = 3, b = 10, c = 2$ (may be implied by correct substitution)","code":"B"},{"id":"p2","marks":1,"description":"Substitutes correctly in

Quadratic Formula Questions

Quadratic Formula Questions

Quadratic Formula Questions

Quadratic Formula Questions