Given that k^2-3k+5=0 then determine the value of k^4-6k^3+9k^2-7.?

Answer 1

using the quadratic formula you'll find that the first equation has complex roots. They are

k = (3 +/- isqrt11)/2

plugging this into that 4th degree equation would be no fun. But there's an easier way. The 4th degree equation is similar to the square of the quadratic, which equals zero.

like so:

k^4 - 6k^3 + 9k^2 - 7
= k^4 - 6k^3 + 19k^2 - 30k + 25 - 10k^2+30k - 32
=(k^2 -3k +5)^2 + 30k - 10k^2 - 32
= 30k - 10k^2 - 32.

Now this is a bit cleaner for figuring complex.

first for k = (3 + isqrt11)/2,
30(3 + isqrt11)/2 - 10((3 + isqrt11)/2)^2 - 32
= 15(3 + isqrt11) - 10(9 + 6isqrt11 -11)/4 - 32
= 45 + 15isqrt11 - 5(3isqrt11 - 1) - 32
= 45 + 15isqrt11 - 15isqrt11 + 5 -32
= 18 shazzam!

and now
30(3 - isqrt11)/2 -10((3-isqrt11)/2)^2 -32
= 45 - 15isqrt11 -10(9 - 6isqrt11 -11)/4 - 32
= 45 - 15isqrt11 -5(-1 - 3isqrt11) - 32
= 45 + 5 - 32 = 18 shazzam!

shazzam! sounds better than QED this isn't a proof just working a problem so I can say shazzam! ain't it cool how both complex roots produce the same result?