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 (2) Given that x² + 8k + k = 0 has equal roots, (b) find the value of k - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1

Given that the equation x² + 8kx + k = 0 has equal roots, we know that the discriminant must be zero. The discriminant Δ is given by:

$Δ = b² - 4ac$

In this case, a = 1, b = 8k, and c = k:

$Δ = (8k)² - 4(1)(k)$

Setting the discriminant to zero:

$(8k)² - 4k = 0$

This simplifies to:

$64k² - 4k = 0$

Factoring out 4k gives:

$4k(16k - 1) = 0$

For k to be non-zero, we take:

16k - 1 = 0 \\ k = rac{1}{16}

Substituting k = (\frac{1}{16}) into our first equation:

$2x + y = 1$

Rearranging this gives:

$y = 1 - 2x$

Now substituting k into the second equation:

$x² - 4(\frac{1}{16})y + 5(\frac{1}{16}) = 0$

This simplifies to:

$x² - \frac{1}{4}y + \frac{5}{16} = 0$

Substituting for y:

$x² - \frac{1}{4}(1 - 2x) + \frac{5}{16} = 0$

This leads to:

$x² - \frac{1}{4} + \frac{1}{2}x + \frac{5}{16} = 0$

Combining terms yields:

$x² + \frac{1}{2}x + \frac{1}{16} = 0$

Factoring gives:

$(x + \frac{1}{4})² = 0$

Thus, the roots yield:

$x = -\frac{1}{4}$

Substituting back to find y:

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

Therefore, the solution of the simultaneous equations is:

$\boxed{(x, y) = (-\frac{1}{4}, \frac{3}{2})}$