Higher November 2024 Paper 3 Q20
20 Solve the simultaneous equations
\(y^2 = 3x^2 + 4\)
\(y + 2x = 7\)
Give your solutions correct to 3 significant figures. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x = 26.3\ldots\), \(y = -45.6\ldots\) and \(x = 1.71\ldots\), \(y = 3.57\ldots\) | M1 | for correct substitution for \(y^2\) or \(x^2\), eg \((7 - 2x)^2 = 3x^2 + 4\) OR for correct rearrangement and expansion of \((7 - 2x)^2\) to obtain 4 terms with all correct without considering signs or for 3 terms out of 4 correct with correct signs and substitution eg \((7 - 2x)^2 = 49 - 14x - 14x + 4x^2\) and \(49 - 14x - 14x + 4x^2 = 3x^2 + 4\) |
| M1 | for method to write a correct simplified equation eg \(x^2 - 28x + 45\ (= 0)\) | |
| M1 | for a method to solve a correct quadratic eg \(\dfrac{28 \pm \sqrt{(-28)^2 - 4 \times 1 \times 45}}{2 \times 1}\) or \(\dfrac{28 \pm \sqrt{604}}{2}\) or \(14 \pm \sqrt{151}\) or \((x - 14)^2 - 14^2 + 45 = 0\) oe | |
| A1 | \(x = 26.2\) to 26.3, \(y = -45.6\) to \(-45.5\) and \(x = 1.7\) to 1.712, \(y = 3.5\) to 3.6 |
Additional guidance
NB \(49 - 28x\) or \(-28x + 4x^2\) can be considered 3 terms out of 4 correct with correct signs
The quadratic does not have to equal 0, ie accept \(x^2 - 28x = -45\)
Can be implied by both \(x\) values correct or both \(y\) values correct
Answers must be correctly paired
(May be in the body of the working)
If answers are given in the range in working and then rounded incorrectly award full marks