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wolfgirl
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Solve the equation. Express the solution as a radical in the simplest form.
 3 years ago
 3 years ago
wolfgirl Group Title
Solve the equation. Express the solution as a radical in the simplest form.
 3 years ago
 3 years ago

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wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
dw:1314895631260:dw
 3 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.2
16(x+3)=x2 x2 16x 48=0
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
do you mean x^2?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
square both sides.
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
Just make sure that whatever solutions you find are valid for the original question.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
(x12)(x4)
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
how did you get that ?
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
I have no idea, let me try it again
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
It's : x^216x48=0 not: x^216x+48=0
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
did I get the answer right though?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
You 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.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
ok i'll try that
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
can you show how you got that ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
\[ax^2+bx+c=0\] \[x=\frac{b \pm \sqrt{b^24ac}}{2a}\] What values did you use for: a=? b=? c=?
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
sorry my comp was dissconnected again
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
ok  so are you ready to continue solving the above ?
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
yeah, I got 4,10
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
what values did you use for a=? b=? c=? when using the quadratic formula ?
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
a=1 b=16 c=48
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
right. can you show your full calculation ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
first, can you calculate this part ? b^24*a*c That's the part inside the square root.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
dw:1314900092634:dwdw:1314899842897:dw
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
the top drawing is supposed to be on bottom
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
I'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, \[x=\frac{b \pm \sqrt{\Delta}}{2a}=\frac{(16) \pm \sqrt{448}}{2(1)}=\frac{16 \pm \sqrt{7*8^2}}{2}\] \[=\frac{16 \pm 8\sqrt{7}}{2}=8 \pm 4\sqrt{7}\]
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
let me know if the above is clear.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
i'm trying to understand it
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
I don't get it
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
can you tell me where it becomes unclear ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
you basically did the same thing in the drawings you posted  just the negative/positive signs got messed up a bit.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
oh I get it, thanks
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
we're not done yet though ... so if anything above is unclear, let me know before we go on. :)
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
ok, explain the third line
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
do you mean the part where the number 8 suddenly pops out from under the square root ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
oh  this ? 448 = 7*64 = 7*8*8 = 7*8^2
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
ok but we're still not done ...
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
The quadratic equation gave us two solutions for x: \[x=8+4\sqrt{7}\] or \[x=84\sqrt{7}\] Right ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
Can a negative number have a real square root ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
so that means that since (x+3) is inside the square root, we cannot allow (x+3) to be negative  right ?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
is tht the answer?
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
yes it is.
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
ok thanks :)
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
\[16(x + 3) = x^2\] \[x^2  16x  48= 0 \]
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
I did squaring ....
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Now I will find Solutions
 3 years ago

Ishaan94 Group TitleBest 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
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Now when we put the solution in Original function
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
\[4(\sqrt{x+3}) = x \]I mean try putting \(4(2 \sqrt7)\)
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
You will get  in root which is not possible that is why only one solution is possible
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
yes, but plugging in that value into (x+3), it is not negative: (x+3) = (4*(2sqrt(7)))+3 = 0.41...
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Yeah You are right ...I did 3 in 8 my mistake..hmm wolfram is wrong
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Let me check wolfram then
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=4%28sqrt%283%2B8++4sqrt7%29%29%3D8++4sqrt7
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Check this out
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
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: 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
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Something wrong with Wolfram I guess
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
hhhmmmm... http://www.wolframalpha.com/input/?i=4*sqrt%28%2884*sqrt%287%29%29%2B3%29%3D%2884*sqrt%287%29%29
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
A result of squaring both sides ...
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
wolfram is right.
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=sqrt%2811+4sqrt7%29*4 yeah you got it squaring removes the minus sign
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Lol I tried to prove that wolfram can be wrong time to time ...but it's always right..
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
yeah, question is  does the teacher expect wolfgirl to take the two values from the quadratic equation and validate them in the original question.
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
coz only one of them is valid.
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
well I don't think teacher would expect that ...hmm maybe they allow calculators
 3 years ago

fiddlearound Group TitleBest ResponseYou've already chosen the best response.1
ok  we can ask wolfgirl what they usually do in class. Maybe she's related to wolfram.
 3 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
Lol ... Good Job though ...really nice
 3 years ago

wolfgirl Group TitleBest ResponseYou've already chosen the best response.0
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
 3 years ago
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