anonymous
  • anonymous
Solve the equation. Express the solution as a radical in the simplest form.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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jhonyy9
  • jhonyy9
where is ?
anonymous
  • anonymous
|dw:1314895631260:dw|
jhonyy9
  • jhonyy9
16(x+3)=x2 x2 -16x -48=0

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anonymous
  • anonymous
do you mean x^2?
anonymous
  • anonymous
square both sides.
jhonyy9
  • jhonyy9
yes
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Just make sure that whatever solutions you find are valid for the original question.
anonymous
  • anonymous
ok
anonymous
  • anonymous
(x-12)(x-4)
anonymous
  • anonymous
how did you get that ?
anonymous
  • anonymous
I have no idea, let me try it again
anonymous
  • anonymous
It's : x^2-16x-48=0 not: x^2-16x+48=0
anonymous
  • anonymous
ok
anonymous
  • anonymous
I got tht
anonymous
  • anonymous
did I get the answer right though?
anonymous
  • anonymous
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.
anonymous
  • anonymous
ok i'll try that
anonymous
  • anonymous
24,-40
anonymous
  • anonymous
can you show how you got that ?
anonymous
  • anonymous
\[ax^2+bx+c=0\] \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] What values did you use for: a=? b=? c=?
anonymous
  • anonymous
sorry my comp was dissconnected again
anonymous
  • anonymous
ok - so are you ready to continue solving the above ?
anonymous
  • anonymous
yeah, I got -4,-10
anonymous
  • anonymous
what values did you use for a=? b=? c=? when using the quadratic formula ?
anonymous
  • anonymous
a=1 b=-16 c=-48
anonymous
  • anonymous
right. can you show your full calculation ?
anonymous
  • anonymous
first, can you calculate this part ? b^2-4*a*c That's the part inside the square root.
anonymous
  • anonymous
|dw:1314900092634:dw||dw:1314899842897:dw|
anonymous
  • anonymous
the top drawing is supposed to be on bottom
anonymous
  • anonymous
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, \[x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\] \[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]
anonymous
  • anonymous
let me know if the above is clear.
anonymous
  • anonymous
i'm trying to understand it
anonymous
  • anonymous
I don't get it
anonymous
  • anonymous
can you tell me where it becomes unclear ?
anonymous
  • anonymous
you basically did the same thing in the drawings you posted - just the negative/positive signs got messed up a bit.
anonymous
  • anonymous
oh I get it, thanks
anonymous
  • anonymous
we're not done yet though ... so if anything above is unclear, let me know before we go on. :)
anonymous
  • anonymous
ok, explain the third line
anonymous
  • anonymous
do you mean the part where the number 8 suddenly pops out from under the square root ?
anonymous
  • anonymous
7*8^2
anonymous
  • anonymous
oh - this ? 448 = 7*64 = 7*8*8 = 7*8^2
anonymous
  • anonymous
yeah
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
ok but we're still not done ...
anonymous
  • anonymous
The quadratic equation gave us two solutions for x: \[x=8+4\sqrt{7}\] or \[x=8-4\sqrt{7}\] Right ?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Can a negative number have a real square root ?
anonymous
  • anonymous
no
anonymous
  • anonymous
so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative - right ?
anonymous
  • anonymous
right
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
positive
anonymous
  • anonymous
is tht the answer?
anonymous
  • anonymous
yes it is.
anonymous
  • anonymous
ok thanks :)
anonymous
  • anonymous
\[16(x + 3) = x^2\] \[x^2 - 16x - 48= 0 \]
anonymous
  • anonymous
I did squaring ....
anonymous
  • anonymous
Now I will find Solutions
anonymous
  • anonymous
\[x = 4(2 +\sqrt7)\] \[x = 4(2 - \sqrt7)\] http://www.wolframalpha.com/input/?i=x^2+-+16x+-+48+%3D+0
anonymous
  • anonymous
Now when we put the solution in Original function
anonymous
  • anonymous
\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 -\sqrt7)\)
anonymous
  • anonymous
You will get - in root which is not possible that is why only one solution is possible
anonymous
  • anonymous
yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2-sqrt(7)))+3 = 0.41...
anonymous
  • anonymous
Yeah You are right ...I did -3 in 8 my mistake..hmm wolfram is wrong
anonymous
  • anonymous
Let me check wolfram then
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8+-+4sqrt7%29%29%3D8+-+4sqrt7
anonymous
  • anonymous
Check this out
anonymous
  • anonymous
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
anonymous
  • anonymous
Yup
anonymous
  • anonymous
Something wrong with Wolfram I guess
anonymous
  • anonymous
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
anonymous
  • anonymous
A result of squaring both sides ...
anonymous
  • anonymous
wolfram is right.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=sqrt%2811+-4sqrt7%29*4 yeah you got it squaring removes the minus sign
anonymous
  • anonymous
Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..
anonymous
  • anonymous
yeah, question is - does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.
anonymous
  • anonymous
coz only one of them is valid.
anonymous
  • anonymous
well I don't think teacher would expect that ...hmm maybe they allow calculators
anonymous
  • anonymous
ok -- we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.
anonymous
  • anonymous
Lol ... Good Job though ...really nice
anonymous
  • anonymous
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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