6. The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 1

To find the points of intersection, set the equations of C and l equal to each other:

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

Multiplying through by $x$ to eliminate the fraction yields:

$3 = 2x^2 + 5x$

Rearranging gives:

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

Using the quadratic formula, where $a = 2$, $b = 5$, and $c = -3$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$

Thus, we find:

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

Now substitute these x-values back into either original equation to find their corresponding y-values:

For $x = \frac{1}{2}$: $y = 2(\frac{1}{2}) + 5 = 6$

For $x = -3$: $y = 2(-3) + 5 = -1$

The coordinates of the points of intersection are therefore:

• $(\frac{1}{2}, 6)$
• $(-3, -1)$