Question

answer the following

Answer 1

$\sqrt{2} (x + p) = 0 \\ x = - 2 \\ \sqrt{2} ( - 2 + p) = 0 \\ p = 2\\ p {x}^{2} + kx + 2 \sqrt{2} \\ 2 {x}^{2} + kx + 2 \sqrt{2} \\ x = - 2 \\ 2 {( - 2)}^{2} + k( - 2) + 2 \sqrt{2} = 0 \\ 8 - 2k + 2 \sqrt{2} = 0 \\ 2k = 8 + 2 \sqrt{2} \\ k = 4 + \sqrt{2}$

Answer 2

x = -2 is root of √2(x+p) = 0
substituting value of x:
√2(p-2) = 0
√2p - 2√2 = 0
√2p = 2√2
p = 2 ___ (i)

-2 is zero of px² + kx + 2√2
therefore 2(-2)² + k(-2) + 2√2 = 0
8 - 2k + 2√2 = 0
4 - k + √2 = 0
k = 4 + √2