Given the simultaneous equations 2x + y = 1 x² - 4ky + 5k = 0 where k is a non zero constant, (a) show that x² + 8kx + k = 0 (b) Given that x² + 8k + k = 0 has equal roots, (c) Find the value of k - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 1

For the equation to have equal roots, the discriminant must be zero:

1. The discriminant for the quadratic equation $x^2 + 8kx + k = 0$ is given by: $D = (8k)^2 - 4(1)(k)$

2. Setting the discriminant to zero: $(8k)^2 - 4k = 0$

3. This simplifies to: $64k^2 - 4k = 0$

4. Factor out k: $k(64k - 4) = 0$

5. Solving this gives: $k = 0$ or $64k - 4 = 0$ Therefore:

   $$$$

ightarrow k = rac{1}{16}$Since k cannot be zero, we have:$k = rac{1}{16}$$

1. Substitute k = rac{1}{16} into the first equation:

   $2x + y = 1$

2. Substitute k into the second equation:

   $x^2 + 8(\frac{1}{16})x + \frac{1}{16} = 0$

   This simplifies to:

   $x^2 + \frac{1}{2}x + \frac{1}{16} = 0$

3. Now apply the quadratic formula:

   $x = \frac{-b \pm \sqrt{D}}{2a}$ where $D = 0$ reveals a repeated root.

   Thus: $x = -\frac{1}{4}$ Now substitute $x$ back into the first equation:

   $y = 1 - 2(-\frac{1}{4}) = 1 + \frac{1}{2} = \frac{3}{2}$

4. Therefore, the solution to the simultaneous equations is: $x = -\frac{1}{4}$ and $y = \frac{3}{2}$.