Question

Can someone please help with this: Find all real solutions of the equation: 2(c-2)^2-7=9

Answer 1

is this your question??
\[\Large 2(c-2)^2-7=9\]

Answer 2

yes

Answer 3

ok using
\[\Large (a-b)^2=a^2+b^2-2ab\]
expand (c-2)^2 as
\[\Large 2(c^2+4-4c)-7=9\]
now simplify\[\Large \implies (2c^2+8-8c)-7=9\]
\[\Large \implies 2c^2-8c-8=0\]
it is Quadratic Equation

Answer 4

@seattle12345 can you do this now ?

Answer 5

@seattle12345 waiting for your reply!

Answer 6

sorry I was working on it. so we need two factors of -16 whose sum is -8 right?

Answer 7

ok i think you should use Quadratic Equation these cannot be factored.
are you familiar with quadratic Equation ?

Answer 8

no we have not gone over that yet

Answer 9

completing squares???
you familiar with completing squares?

Answer 10

haven't covered that either

Answer 11

ok
original equation is
\[\Large 2(c-2)^2-7=9\]
add seven to both sides
\[\Large \implies 2(c-2)^2-7+7=9+7\]
\[\Large \implies 2(c-2)^2=16\]
divide both sides by 2
\[\Large \implies \frac{2(c-2)^2}{2}=\frac{16}{2}\]
\[\Large \implies (c-2)^2=8\]
taking square root of both sides
\[\Large \implies \sqrt{(c-2)^2}=\sqrt{8}\]
\[\Large \implies (c-2)=\pm \sqrt{8}\]

can you do this now ??

Answer 12

@seattle12345

Answer 13

yes! thank you!

Answer 14

you are welcome:)