Solve the simultaneous equations $$y - 2x - 4 = 0$$ $$4x^2 + y^2 + 20x = 0$$ - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 1

Now substitute $y = 2x + 4$ into the second equation: $4x^2 + (2x + 4)^2 + 20x = 0$.

Expanding the equation: $4x^2 + (4x^2 + 16x + 16) + 20x = 0$ Combine like terms: $8x^2 + 36x + 16 = 0$.

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 8$, $b = 36$, and $c = 16$:

First calculate the discriminant: $b^2 - 4ac = 36^2 - 4 \cdot 8 \cdot 16 = 1296 - 512 = 784$. Now plug the values into the quadratic formula:

$x = \frac{-36 \pm \sqrt{784}}{2 \cdot 8} = \frac{-36 \pm 28}{16}$. This gives us two solutions for x:

1. $x = \frac{-8}{16} = -0.5$
2. $x = \frac{-64}{16} = -4$.

Now, we can find the corresponding values of y for these x values:

1. When $x = -0.5$: $y = 2(-0.5) + 4 = 3$
2. When $x = -4$: $y = 2(-4) + 4 = -4$.

Thus, the solutions to the simultaneous equations are:

1. $(-0.5, 3)$
2. $(-4, -4)$.