| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Substitution to solve disguised quadratic |
| Difficulty | Moderate -0.5 This is a straightforward substitution question requiring students to manipulate fractional indices and solve a quadratic. Part (i) is guided algebra showing the substitution works, and part (ii) involves factorising/using the quadratic formula then reversing the substitution. While it requires multiple steps, the techniques are routine C1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02f Solve quadratic equations: including in a function of unknown |
\begin{enumerate}
\item (i) Given that $y = x ^ { \frac { 1 } { 3 } }$, show that the equation
\end{enumerate}
$$2 x ^ { \frac { 1 } { 3 } } + 3 x ^ { - \frac { 1 } { 3 } } = 7$$
can be rewritten as
$$2 y ^ { 2 } - 7 y + 3 = 0 .$$
(ii) Hence, solve the equation
$$2 x ^ { \frac { 1 } { 3 } } + 3 x ^ { - \frac { 1 } { 3 } } = 7$$
\hfill \mbox{\textit{OCR C1 Q6 [7]}}