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anonymous
 5 years ago
Solve the equation. Express the solution as a radical in the simplest form.
anonymous
 5 years ago
Solve the equation. Express the solution as a radical in the simplest form.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1314895631260:dw

jhonyy9
 5 years ago
Best ResponseYou've already chosen the best response.216(x+3)=x2 x2 16x 48=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just make sure that whatever solutions you find are valid for the original question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have no idea, let me try it again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's : x^216x48=0 not: x^216x+48=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did I get the answer right though?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You got this right: x^216x+48 = (x12)(x4) But we need to solve this: x^216x48=0 Probably easiest to just use the quadratic formula.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you show how you got that ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ax^2+bx+c=0\] \[x=\frac{b \pm \sqrt{b^24ac}}{2a}\] What values did you use for: a=? b=? c=?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry my comp was dissconnected again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok  so are you ready to continue solving the above ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what values did you use for a=? b=? c=? when using the quadratic formula ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right. can you show your full calculation ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first, can you calculate this part ? b^24*a*c That's the part inside the square root.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1314900092634:dwdw:1314899842897:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the top drawing is supposed to be on bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll start with this guy under the square root. He is a special guy and even has a name: The discriminant. \[\Delta=b^24ac\] \[\Delta=b^24ac=(16)^24(1)(48)=256+192=448\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0let me know if the above is clear.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm trying to understand it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you tell me where it becomes unclear ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you basically did the same thing in the drawings you posted  just the negative/positive signs got messed up a bit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we're not done yet though ... so if anything above is unclear, let me know before we go on. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, explain the third line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mean the part where the number 8 suddenly pops out from under the square root ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh  this ? 448 = 7*64 = 7*8*8 = 7*8^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok but we're still not done ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The quadratic equation gave us two solutions for x: \[x=8+4\sqrt{7}\] or \[x=84\sqrt{7}\] Right ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0Can a negative number have a real square root ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative  right ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[16(x + 3) = x^2\] \[x^2  16x  48= 0 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now I will find Solutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x = 4(2 +\sqrt7)\] \[x = 4(2  \sqrt7)\] http://www.wolframalpha.com/input/?i=x^2++16x++48+%3D+0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now when we put the solution in Original function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 \sqrt7)\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You will get  in root which is not possible that is why only one solution is possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2sqrt(7)))+3 = 0.41...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah You are right ...I did 3 in 8 my mistake..hmm wolfram is wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me check wolfram then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8++4sqrt7%29%29%3D8++4sqrt7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Checking: 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: 84*sqrt(7)  Wolfram says false: http://www.wolframalpha.com/input/?i=4*sqrt%28%2884*sqrt%287%29%29%2B3%29%3D%2884*sqrt%287%29%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Something wrong with Wolfram I guess

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hhhmmmm... http://www.wolframalpha.com/input/?i=4*sqrt%28%2884*sqrt%287%29%29%2B3%29%3D%2884*sqrt%287%29%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A result of squaring both sides ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=sqrt%2811+4sqrt7%29*4 yeah you got it squaring removes the minus sign

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, question is  does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0coz only one of them is valid.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well I don't think teacher would expect that ...hmm maybe they allow calculators

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok  we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol ... Good Job though ...really nice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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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