Question

Can someone please check my work? I'll post my work and answer in the comments. I'm not sure if I completed it properly.

Answer 1

Find the solutions of [2x ^{2}-16x+32=0\]

Answer 2

My work:
\[x= 16\pm \frac{\sqrt{16^2}-4(2)(32) }{ 2(2) }\]

\[x= 16\pm \frac{ 256-4(64) }{ 4 }\]

\[x= 16\pm \frac{\sqrt{256 -256}}{ 4 }\]

\[x= 16\pm \sqrt{0}\]

\[x= \frac{ 16+0 }{ 4 } , \frac{ 16-0 }{ 4 } = (0,0)\]

Answer 3

If x = 0 then that should mean there are no solutions, correct?

Answer 4

Because when I check my work I get:
\[2(0)^2-16(0)+32\]

\[0-0+32=0\]

\[32 \neq 0\]

Answer 5

Where did the 0 come from?

\[\large x = \frac{16 \pm 0}{4} \cancel{=} 0\]

Answer 6

When I took

x=16±√0

and separated the addition and subtraction symbols so I would have two equations

\[x= \frac{ 16+0 }{ 4 } , \frac{ 16-0 }{ 4 }\]

Answer 7

Oh wait

Answer 8

I see what I did wrong, it would \[\frac{ 16 }{ 4 }\] regardless of addition or subtraction

Answer 9

There ya go!

Answer 10

Thank you so much! I can't believe I missed something so simple.

Answer 11

Lol it happens dont worry about it!