A-LEVEL•Mathematics•Quadratics•Question 9

Solution

🟡 Exam-style · [4 marks] Solve the equation $2x - 7\sqrt{x} + 3 = 0$.

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Let $u = \sqrt{x}$ , so $x = u^2$ :

\begin{aligned} 2u^2 - 7u + 3 &= 0 \\ (2u - 1)(u - 3) &= 0 \\ u = \tfrac12 \quad&\text{or}\quad u = 3 \end{aligned}

Both values are positive, so both are valid. $\sqrt{x} = \tfrac12$ gives $x = \tfrac14$ , and $\sqrt{x} = 3$ gives $x = 9$ .

Check $x = 9$ : $18 - 21 + 3 = 0$ ✓ Check $x = \tfrac14$ : $\tfrac12 - \tfrac72 + 3 = 0$ ✓

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