6. (a) By eliminating y from the equations y = x - 4, 2x^2 - xy = 8, show that x^2 + 4x - 8 = 0 - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 1

To solve the simultaneous equations, we can use the previously derived quadratic equation:

$x^2 + 4x - 8 = 0$

We apply the quadratic formula, given by:

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

In our quadratic equation, we have:

• a = 1
• b = 4
• c = -8

Substituting these values into the formula:

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

Simplifying the discriminant:

$= \frac{-4 \pm \sqrt{16 + 32}}{2}$ $= \frac{-4 \pm \sqrt{48}}{2}$ $= \frac{-4 \pm 4\sqrt{3}}{2}$ $= -2 \pm 2\sqrt{3}$

Thus, we have two solutions for x:

$x = -2 + 2\sqrt{3}$ $x = -2 - 2\sqrt{3}$

Now, we substitute back to find y. Using the equation:

$y = x - 4$

1. For $x = -2 + 2\sqrt{3}$:

   $y = (-2 + 2\sqrt{3}) - 4 = -6 + 2\sqrt{3}$

2. For $x = -2 - 2\sqrt{3}$:

   $y = (-2 - 2\sqrt{3}) - 4 = -6 - 2\sqrt{3}$

Finally, the solutions to the simultaneous equations are:

$x = -2 + 2\sqrt{3}, y = -6 + 2\sqrt{3}$ $x = -2 - 2\sqrt{3}, y = -6 - 2\sqrt{3}$

These results are in the form a ± b√3, where a and b are integers.