Figure 4 shows a sketch of part of the curve $C_1$ with equation $y = 2x^2 + 10 \quad x > 0$ and part of the curve $C_2$ with equation $y = 42x - 15x^2 - 7 \quad x > 0$ (a) Verify that the curves intersect at $x = \frac{1}{2}$ (b) The curves intersect again at the point $P$ - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 1

To find the second intersection point $P$, we need to set the equations equal:

$2x^2 + 10 = 42x - 15x^2 - 7$

Rearranging gives:

$2x^2 + 15x^2 - 42x + 10 + 7 = 0$

This simplifies to:

$17x^2 - 42x + 17 = 0$

Now we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 17$, $b = -42$, and $c = 17$. Plugging in:

$x = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 17 \cdot 17}}{2 \cdot 17}$

Calculating the discriminant:

$(-42)^2 = 1764 \\ 4 \cdot 17 \cdot 17 = 1156 \\ 1764 - 1156 = 608$

Thus:

$x = \frac{42 \pm \sqrt{608}}{34} = \frac{42 \pm 4\sqrt{38}}{34} = \frac{21 \pm 2\sqrt{38}}{17}$

Since we already know $x = \frac{1}{2}$, the second value must be:

$x = \frac{21 - 2\sqrt{38}}{17}$

Finally, we conclude the second intersection point occurs where:

$x_P = \frac{21 - 2\sqrt{38}}{17}$.