Question

Find K

\[(2k-2)^2-4(-4k+36)=0\]

Answer 1

It's OK. I solved it!!

Answer 2

We have
\[(2k-2)^2-4(-4k+36)=0\]
We could expand and solve for k. But we'll try a different method
Let's rewrite the equation
\[(2k-2)^2-4*(-2)(2k-18)=0\]
or
[(2k-2)^2+8(2k-2-16)=0\]
Let's substitute 2k-2=r
\[r^2+8(r-16)=0\]
now
\[r^2+8r-16=0\]
Let's add 16 and subtract 16
\[r^2+8r+16-32=0\]
now
\[(r+4)^2-32=0\]
or
\[r+4=\pm \sqrt {32}\]
r=2k-2
\[2k+2=\pm 4\sqrt {2}\]
we get
\[k=-1\pm 2\sqrt{2}\]

Answer 3

That's not the right answer, but thanks for trying anway. I have the answer :D

Answer 4

What answer you have? I'll recheck

Answer 5

k= -7 or 5

Answer 6

Got my mistake. :P

Answer 7

great :D

Answer 8

\[r^2+8(r-16)=0\]
\(r^2+8r-128=0\) (this is where I made mistake earlier)
add and subtract 16
\[r^2+8r+16-144=0\]
\[(r+4)^2=144\]
\[r+4=\pm 12\]
r=2k-2
\[2k-2=\pm 12\]
\[2k=2\pm 12\]
\[k=1 \pm 6\]
\[k=-5 or 7\]
:D

Answer 9

Yeah, got it :D