Question

i write the wrong question 2(x-2)^2-6
find allints in exact form

Answer 1

very similar to the last one

Answer 2

replace x by 0 to get the y intercept
set y = 0 and solve for x to get the x intercepts

Answer 3

y=2

Answer 4

but i need help with the x's

Answer 5

yes y = 2

Answer 6

\[2(x-2)^2-6=0\]
\[2(x-2)^2=6\]
\[(x-2)^2=3\]
\[x-2=\pm\sqrt{3}\]
\[x=2\pm\sqrt{3}\]

Answer 7

make sure you follow all the steps. they are all there i skipped nothing

Answer 8

so you dont have to expand the brackets??

Answer 9

oh heck no

Answer 10

it is just in the form i want it. check the steps.
if you expand you have way more work to do after you expand, so you would be working against yourself

Answer 11

yeh cos i was expanding and going WTF this clears it up just 1 question why is it \[\pm\] not just +

Answer 12

because if you have
\[x^2=25\] you have two choices for x
\[x=5\] or
\[x=-5\] so it is an abbreviation for the two choices

Answer 13

ohn k thanks'

Answer 14

\[(x-2)^2=25\]
\[x-2=5\] or
\[x-2=-5\]
\[x=7\] or
\[x=-3\]

Answer 15

\[(x-2)^2=3\]
\[x-2=\sqrt{3}\] or
\[x-2=-\sqrt{3}\]
\[x=2+\sqrt{3}\] or
\[x=2-\sqrt{3}\]