wolfgirl
Solve the equation. Express the solution as a radical in the simplest form.
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jhonyy9
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where is ?
wolfgirl
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|dw:1314895631260:dw|
jhonyy9
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16(x+3)=x2
x2 -16x -48=0
wolfgirl
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do you mean x^2?
fiddlearound
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square both sides.
jhonyy9
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yes
wolfgirl
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oh ok
fiddlearound
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Just make sure that whatever solutions you find are valid for the original question.
wolfgirl
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ok
wolfgirl
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(x-12)(x-4)
fiddlearound
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how did you get that ?
wolfgirl
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I have no idea, let me try it again
fiddlearound
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It's :
x^2-16x-48=0
not:
x^2-16x+48=0
wolfgirl
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ok
wolfgirl
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I got tht
wolfgirl
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did I get the answer right though?
fiddlearound
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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.
wolfgirl
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ok i'll try that
wolfgirl
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24,-40
fiddlearound
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can you show how you got that ?
fiddlearound
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\[ax^2+bx+c=0\]
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
What values did you use for:
a=?
b=?
c=?
wolfgirl
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sorry my comp was dissconnected again
fiddlearound
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ok - so are you ready to continue solving the above ?
wolfgirl
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yeah, I got -4,-10
fiddlearound
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what values did you use for
a=?
b=?
c=?
when using the quadratic formula ?
wolfgirl
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a=1 b=-16 c=-48
fiddlearound
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right. can you show your full calculation ?
fiddlearound
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first, can you calculate this part ?
b^2-4*a*c
That's the part inside the square root.
wolfgirl
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|dw:1314900092634:dw||dw:1314899842897:dw|
wolfgirl
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the top drawing is supposed to be on bottom
fiddlearound
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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,
\[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\]
\[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]
fiddlearound
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let me know if the above is clear.
wolfgirl
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i'm trying to understand it
wolfgirl
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I don't get it
fiddlearound
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can you tell me where it becomes unclear ?
fiddlearound
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you basically did the same thing in the drawings you posted -
just the negative/positive signs got messed up a bit.
wolfgirl
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oh I get it, thanks
fiddlearound
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we're not done yet though ... so if anything above is unclear, let me know before we go on.
:)
wolfgirl
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ok, explain the third line
fiddlearound
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do you mean the part where the number 8 suddenly pops out from under the square root ?
wolfgirl
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7*8^2
fiddlearound
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oh - this ?
448 = 7*64 = 7*8*8 = 7*8^2
wolfgirl
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yeah
fiddlearound
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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 ?
wolfgirl
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yeah
fiddlearound
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ok but we're still not done ...
fiddlearound
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The quadratic equation gave us two solutions for x:
\[x=8+4\sqrt{7}\]
or
\[x=8-4\sqrt{7}\]
Right ?
wolfgirl
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yeah
fiddlearound
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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 ?
wolfgirl
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yeah
fiddlearound
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Can a negative number have a real square root ?
wolfgirl
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no
fiddlearound
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so that means that since (x+3) is inside the square root,
we cannot allow (x+3) to be negative - right ?
wolfgirl
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right
fiddlearound
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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 ?
wolfgirl
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positive
wolfgirl
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is tht the answer?
fiddlearound
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yes it is.
wolfgirl
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ok thanks :)
Ishaan94
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\[16(x + 3) = x^2\]
\[x^2 - 16x - 48= 0 \]
Ishaan94
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I did squaring ....
Ishaan94
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Now I will find Solutions
Ishaan94
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Now when we put the solution in Original function
Ishaan94
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\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)
Ishaan94
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You will get - in root which is not possible that is why only one solution is possible
fiddlearound
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yes,
but plugging in that value into (x+3), it is not negative:
(x+3) = (4*(2-sqrt(7)))+3 = 0.41...
Ishaan94
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Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong
Ishaan94
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Let me check wolfram then
Ishaan94
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Check this out
Ishaan94
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Yup
Ishaan94
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Something wrong with Wolfram I guess
fiddlearound
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A result of squaring both sides ...
fiddlearound
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wolfram is right.
Ishaan94
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Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..
fiddlearound
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yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.
fiddlearound
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coz only one of them is valid.
Ishaan94
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well I don't think teacher would expect that ...hmm maybe they allow calculators
fiddlearound
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ok -- we can ask wolfgirl what they usually do in class.
Maybe she's related to wolfram.
Ishaan94
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Lol ...
Good Job though ...really nice
wolfgirl
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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