The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 1

To find the points of intersection:

1. Set the equations equal:

   $\frac{3}{x} = 2x + 5$

   Multiply both sides by $x$ (noting $x \neq 0$):

   $3 = 2x^2 + 5x$

   Rearranging gives:

   $2x^2 + 5x - 3 = 0$

2. Use the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ with $a = 2, b = 5, c = -3$:

   $x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-3)}}{2 \times 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$

   Thus, we have two solutions:

   • $x = \frac{2}{4} = \frac{1}{2}$
   • $x = \frac{-12}{4} = -3$

3. Finding y-values:

   For $x = \frac{1}{2}$:

   $y = 2\left(\frac{1}{2}\right) + 5 = 1 + 5 = 6$

   For $x = -3$:

   $y = 2(-3) + 5 = -6 + 5 = -1$

So the points of intersection are:

• $(\frac{1}{2}, 6)$
• $(-3, -1)$