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Solve the equation. Express the solution as a radical in the simplest form.

Mathematics
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where is ?
|dw:1314895631260:dw|
16(x+3)=x2 x2 -16x -48=0

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Other answers:

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?
You got this right: x^2-16x+48 = (x-12)(x-4) But we need to solve this: x^2-16x-48=0 Probably easiest to just use the quadratic formula.
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
I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant. \[\Delta=b^2-4ac\] \[\Delta=b^2-4ac=(-16)^2-4(1)(-48)=256+192=448\] so, \[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\] \[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]
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 ?
you basically did the same thing in the drawings you posted - just the negative/positive signs got messed up a bit.
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
the reason I wrote it that way, is because the question wants us to: "Express the solution as a radical in the simplest form" Which means that we need to take everything we can out from under the square root. \[\sqrt{448}=\sqrt{7*64}=\sqrt{7*8^2}=\sqrt{7}\sqrt{8^2}=\sqrt{7}*8=8\sqrt{7}\] Is that cleat now ?
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
but if you look at the original question, which was: \[4\sqrt{x+3}=x\] Notice the there is a square root there, and (x+3) is inside the square root. Right ?
yeah
Can a negative number have a real square root ?
no
so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative - right ?
right
so we need to check the two possible values we got for x from the quadratic equation, and check that they don't make the number (x+3) a negative number. One of the numbers is easy to check, try checking this one first: \[x=8+4\sqrt{7}\] When you plug the above value into: \[(x+3)\] Do you get a negative value, or not ?
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
Checking: 8+4*sqrt(7) - Wolfram says true: http://www.wolframalpha.com/input/?i=4*sqrt%28%288%2B4*sqrt%287%29%29%2B3%29%3D%288%2B4*sqrt%287%29%29 Checking: 8-4*sqrt(7) - Wolfram says false: http://www.wolframalpha.com/input/?i=4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29
Yup
Something wrong with Wolfram I guess
hhhmmmm... http://www.wolframalpha.com/input/?i=-4*sqrt%28%288-4*sqrt%287%29%29%2B3%29%3D%288-4*sqrt%287%29%29
A result of squaring both sides ...
wolfram is right.
http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4 yeah you got it squaring removes the minus sign
Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..
yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.
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
I'm not related to wolfram lol. But so far I am clueless as to what they do in class. I do Ashworth online school if that helps any

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