Question

help help will medal

Answer 1

HI!!

Answer 2

Hi darling xx

Answer 3

does "enter the quadratic equation in standard form mean make it look like \[ax^2+bx+c=0\]?

Answer 4

yes

Answer 5

\[2x^2-12+4x=(x-2)^2+2\]

Answer 6

gotta expand on the right, start with \((x+2)(x+2)\) and multiply
you know how to do that ?

Answer 7

Um kind of

Answer 8

hmm
maybe it should be \[(x-2)(x-2)\]

Answer 9

oh yeah that one \[(x-2)(x-2)=x^2-2x-2x+4=x^2-4x+4\]

Answer 10

it is always the case that \((a+b)^2=a^2+2ab+b^2\) so might as well just go from \[(x-2)(x-2)\) to \(x^2-4x+4\)

Answer 11

lol
nice latex
so on the right you get \[x^2-4x+4+2=x^2-4x+6\]

Answer 12

oh okay and then would i move the terms ?

Answer 13

we are looking at \[2x^2-12+4x=(x-2)^2+2\\
2x^2-12+4x=x^2-4x+6\]

Answer 14

subtract \(x^2\) from both sides

Answer 15

left with 2-12+4x=-4x+6?

Answer 16

oh no dear, lets go slow

Answer 17

i know you think that if you take \(x^2\) away from \(2x^2\) you get \(2\) but that is not right
\(x\) is a variable, lets say \(x^2=100\)
what is \(200-100\)?

Answer 18

ojhhhh 1?

Answer 19

its the sq rt?

Answer 20

\(200-100=?\)

Answer 21

100 beautiful =)

Answer 22

zarkam: This problem is not much different from previous ones with which I helped you. The main difference is that you have a more complicated right side. You must multiply out that (x-2)^2. Before doing anything else, look at the right side and combine "like terms."
Then, just as we did before, move all of the terms now on the right side to the left side.

Answer 23

yes darling
and two apples minus one apple is one apple

Answer 24

and two corvettes minus one corvette is one corvette

Answer 25

But would it be the sq rt?

Answer 26

nope, no square roots are involved here

Answer 27

Oh okay

Answer 28

just\[2\heartsuit -1\heartsuit=1\heartsuit\]

Answer 29

It may help you, zarkam, if you write down what you learned from each correctly solved problem. that way, you'd have something to which to refer when you feel stuck on a new problem.

Right: NO square roots here. Expand (x-2)^2 and you'll get a square (the square of x), not a square root.

Answer 30

i.e \[2x^2-x^2=x^2\] not \[2x^2-x^2=2\]

Answer 31

Awe okay :* <3

Answer 32

Oh okay so 2 I get it now

Answer 33

so now were are here \[2x^2-12+4x=(x-2)^2+2\\
2x^2-12+4x=x^2-4x+6\\
x^2-12+4x=-4x+6\]

Answer 34

now add \(4x\) to both sides

Answer 35

x^2-12+8x=6?

Answer 36

right, one more step (actually two)
subtract 6

Answer 37

x^2-18+8x=0?

Answer 38

right, but last step is to put in standard form
but the \(+8x\) before the \(-18\)

Answer 39

So:
x^2+8x-18=0

Answer 40

yes, that should do it ?

Answer 41

Thanks honeysuckle =) <3

Answer 42

your welcome sweet one \[\color\magenta\heartsuit\]