The equation of a curve is $(x + y)^2 = 4y + 2x + 8$ The curve intersects the positive x-axis at the point P - AQA - A-Level Maths Mechanics - Question 12 - 2021 - Paper 1

Next, we can factor or use the quadratic formula:

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

Substituting $a = 1, b = -2, c = -8$:

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$

This results in:

$x = \frac{2 \pm \sqrt{36}}{2}$

Thus:

$x = \frac{2 \pm 6}{2}$

This yields two solutions: $x = 4$ and $x = -2$. Since we consider only the positive x-axis, we take $x = 4$.

To find the gradient, we differentiate the given equation implicitly:

Start from:

$(x + y)^2 = 4y + 2x + 8$

Differentiating both sides will involve using the product rule for the left-hand side:

$2(x + y) \left(1 + \frac{dy}{dx}\right) = 4 \frac{dy}{dx} + 2$

Substituting $x = 4$ and $y = 0$ into the differentiated equation gives:

$2(4 + 0) \left(1 + \frac{dy}{dx}\right) = 4 \frac{dy}{dx} + 2$

This simplifies to:

$8 \left(1 + \frac{dy}{dx}\right) = 4 \frac{dy}{dx} + 2$

Expanding this yields:

$8 + 8 \frac{dy}{dx} = 4 \frac{dy}{dx} + 2$

Rearranging yields:

$8 \frac{dy}{dx} - 4 \frac{dy}{dx} = 2 - 8$

This simplifies to:

$4 \frac{dy}{dx} = -6$

Thus, solving for the gradient gives:

$\frac{dy}{dx} = -\frac{3}{2}$

This is the gradient at point P, showing the required result.