Question

Ok maybe someone can get this one. Find all solutions of the equation and express them in the form
a + bi.
64x^2 + 3 = 16x
x = ____ (positive imaginary part)
x = ____ (negative imaginary part)

Answer 1

The best way to do this problem is to use the quadratic formula. First, put zero on one side and everything else on the other
\[0=64x^2-16x+3\]Plug this into the quadratic formula with a=64, b=-16, c=3.
You end up with \[x=(16\pm \sqrt{-512})/128\]We can remove the negative part from the inside of the radical to get
\[x=(16\pm i \sqrt{512})/128\]The perfect square 256 goes evenly into 512, so we pull that out of the square root as a 16.
\[x=(16\pm 16i \sqrt{2})/128\]Next, we notice that 16 goes evenly into 128.
\[x=(1\pm i \sqrt{2})/8\]Another way to write this is
\[x=1/8 \pm i \sqrt{2}/8\]This yields two solutions
\[x=1/8+(\sqrt{2}/8)i\]\[x=1/8-(\sqrt{2}/8)i\]

Answer 2

so whats the actual answer

Answer 3

The last two lines.

Answer 4

it cant be simplified anymore

Answer 5

Nope. There are two solutions, and each one has a real part and an imaginary part. Your a = 1/8 and your b = sqrt(2)/8. Whenever you have a solution x=a+bi, you also have x=a-bi.

Answer 6

can you look at my last ? i asked too because people tried to explain it and i couldnt get it

Answer 7

sorry thats the last one