Foundation June 2025 Paper 1 Q26
26
Not drawn accurately
The area of the rectangle is 120 cm\(^2\)
Work out the value of \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + 2x - 5x - 10\) | M1 | oe with brackets expanded four terms in any order with three correct from \(x^2 \quad (+)2x \quad -5x \quad -10\) terms may be seen in a grid implied by \(x^2 - 3x + k \quad (k \ne 0)\) or \(ax^2 - 3x - 10 \quad (a \ne 0)\) |
| \(x^2 - 3x - 130\ (= 0)\) | M1dep | oe expression/equation with brackets expanded eg \(x^2 + 2x - 5x - 10 = 120\) |
| For their three-term quadratic, correctly factorises or correctly substitutes into the quadratic formula or correctly completes the square to the form \(x = \ldots\) for their quadratic or \(-10\) and 13 | M1 | do not accept \(x^2 - 3x - 10\ (= 0)\) as their three-term quadratic eg \((x + 10)(x - 13)\ (= 0)\) eg \(\dfrac{--3 \pm \sqrt{(-3)^2 - 4 \times 1 \times -130}}{2 \times 1}\) eg \((x =)\ 1.5 \pm \sqrt{\left(\dfrac{3}{2}\right)^2 + 130}\) |
| 13 | A1 | SC1 31.5 oe |
Additional guidance
The first and third marks may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
SC1 is for using the perimeter
Trial and improvement is 0, 3 (for \(-10\) and 13) or 4 marks