Atroshus

I'm sorry I was posting the answer I found another way as a reply to your answer.

Just found the answer,

Let z=a+bi and \(\bar{z}=a-bi\).

so \(4i\bar{z}=4b+4ia\).

We now find that \(3z+4i\bar{z} = (3a+4b) + (4a+3b)i\)

so \(3a+4b=1\) and \(4a+3b=-8\)

Solving this leads us to a=-5 and b=4

So the answer is \(z=\boxed{-5+4i}\)

omg thx ur a hero :D

not correct

Complete the square

\(f(x) = (x^2+2ax+a^2) + (3a-a^2) = (x+a)^2 + (3a-a^2)\)

So the minimum value is 3a-a^2

so \(3a - a^2 = 2\)

a=1,2

Multiply both sides by (x-4)(2x+1)

then we get (2x + 3)(2x + 1) - (2x - 8)(x - 4) = (x - 4)(2x + 1)

Simplifies to 31x - 25 = 0

so x=25/31