Higher June 2019 Paper 1R Q22

\(2x^2 + 3(2x + 1)^2 = 5\)

---

or

\(2\left(\dfrac{y - 1}{2}\right)^2 + 3y^2 = 5\)

eg \((7x - 1)(x + 1)\) or \((7x - 1)(2x + 2)\)

eg \(\dfrac{-12 \pm \sqrt{12^2 - 4 \times 14 \times -2}}{2 \times 14}\) oe

eg \(14\left(\left(x + \dfrac{3}{7}\right)^2 - \dfrac{9}{49}\right) = 2\) oe (corrected from the printed mark scheme: \(7\left(\left(x + \dfrac{3}{7}\right)^2 - \dfrac{9}{49}\right) = 2\))

---

or

eg \((7y - 9)(y + 1)\)

eg \(\dfrac{2 \pm \sqrt{(-2)^2 - 4 \times 7 \times -9}}{2 \times 7}\) oe

eg \(7\left(\left(y - \dfrac{1}{7}\right)^2 - \dfrac{1}{49}\right) = 9\) oe

\(x = \dfrac{1}{7}\), \(x = -1\) (need both)

---

or

\(y = \dfrac{9}{7}\), \(y = -1\) (need both)

Answer: \(x = \dfrac{1}{7}\), \(y = \dfrac{9}{7}\)

\(x = -1\), \(y = -1\)