Question

for questions 9-10 what are the solutions of the equation
A. c^2-4c=0
B. z^2-4c-27=0

Answer 1

will give medal

Answer 2

\[\huge c^2- 4c=0 \rightarrow c(c-4)= 0\]
\[\huge \rightarrow c =0 or (c-4)= 0\]
\[\huge \rightarrow c =0 or c=4\]
c= 0, 4

Answer 3

ok thanks what about B

Answer 4

\[\huge z^2-4c-27 =0\]
Let us put c=0 , we find;

\[\huge z^2-4\times 0-27 =0 \rightarrow z^2- 0-27 =0\]
\[\huge \rightarrow z^2- 27 =0 \rightarrow z^2 = 27 \rightarrow z= \pm 3\]

Answer 5

i goofed up its z^2-4z-27=0 does that change any thing

Answer 6

At c = 4, we find \[\huge \rightarrow z^2- 4 \times 4-27 =0\]
\[\huge \rightarrow z^2- 16-27 =0\]
\[\huge \rightarrow z^2- 16-27 =0 \rightarrow z^2- 43=0\]
\[\huge \rightarrow z^2= 43 \rightarrow z= \pm \sqrt {43}\]

Answer 7

@handyandy3

Answer 8

my answer choices are
3,9
3,-9
-9,3
-3,-9

Answer 9

@handyandy3

Answer 10

yeah

Answer 11

so are you able to figure it out?

Answer 12

@dpasingh