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where is ?

|dw:1314895631260:dw|

16(x+3)=x2
x2 -16x -48=0

do you mean x^2?

square both sides.

yes

oh ok

Just make sure that whatever solutions you find are valid for the original question.

ok

(x-12)(x-4)

how did you get that ?

I have no idea, let me try it again

It's :
x^2-16x-48=0
not:
x^2-16x+48=0

ok

I got tht

did I get the answer right though?

ok i'll try that

24,-40

can you show how you got that ?

\[ax^2+bx+c=0\]
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
What values did you use for:
a=?
b=?
c=?

sorry my comp was dissconnected again

ok - so are you ready to continue solving the above ?

yeah, I got -4,-10

what values did you use for
a=?
b=?
c=?
when using the quadratic formula ?

a=1 b=-16 c=-48

right. can you show your full calculation ?

first, can you calculate this part ?
b^2-4*a*c
That's the part inside the square root.

|dw:1314900092634:dw||dw:1314899842897:dw|

the top drawing is supposed to be on bottom

let me know if the above is clear.

i'm trying to understand it

I don't get it

can you tell me where it becomes unclear ?

oh I get it, thanks

we're not done yet though ... so if anything above is unclear, let me know before we go on.
:)

ok, explain the third line

do you mean the part where the number 8 suddenly pops out from under the square root ?

7*8^2

oh - this ?
448 = 7*64 = 7*8*8 = 7*8^2

yeah

yeah

ok but we're still not done ...

The quadratic equation gave us two solutions for x:
\[x=8+4\sqrt{7}\]
or
\[x=8-4\sqrt{7}\]
Right ?

yeah

yeah

Can a negative number have a real square root ?

no

right

positive

is tht the answer?

yes it is.

ok thanks :)

\[16(x + 3) = x^2\]
\[x^2 - 16x - 48= 0 \]

I did squaring ....

Now I will find Solutions

\[x = 4(2 +\sqrt7)\]
\[x = 4(2 - \sqrt7)\]
http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0

Now when we put the solution in Original function

\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)

You will get - in root which is not possible that is why only one solution is possible

yes,
but plugging in that value into (x+3), it is not negative:
(x+3) = (4*(2-sqrt(7)))+3 = 0.41...

Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong

Let me check wolfram then

http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8+-+4sqrt7%29%29%3D8+-+4sqrt7

Check this out

Yup

Something wrong with Wolfram I guess

A result of squaring both sides ...

wolfram is right.

Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

coz only one of them is valid.

well I don't think teacher would expect that ...hmm maybe they allow calculators

ok -- we can ask wolfgirl what they usually do in class.
Maybe she's related to wolfram.

Lol ...
Good Job though ...really nice