anonymous
  • anonymous
Solve and determine if a solution is extraneous. Can someone please explain.Thanks! In progress...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sqrt{x+16}+\sqrt{x=8}\]
whpalmer4
  • whpalmer4
I think that is probably \[\sqrt{x+16} + \sqrt{x} = 8\]Correct?
anonymous
  • anonymous
yes correct sorry

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Jhannybean
  • Jhannybean
so solve for x we could square each side.
anonymous
  • anonymous
yes that i kind of understand after squared then what?
anonymous
  • anonymous
\[\left( \sqrt{x+16}+\sqrt{x=8}\right)^2\]
Loser66
  • Loser66
I am not sure whether I am right or wrong, just try. watch me, please let x = a^2 . I try to form the form of |a +4i| = sqrt (a^2 +4^2| and the second one is sqrt(a^2) = |a|
Loser66
  • Loser66
then |a+4i|+|a|= 8 solve for a, plug back to solve for x
Jhannybean
  • Jhannybean
\[\large \sqrt{x+16} + \sqrt{x} = 8\] square both sides \[\large (\sqrt{x+16} + \sqrt{x})^2 = (8)^2\] expand \[\large (\sqrt{x+16} +\sqrt{x})(\sqrt{x+16}+\sqrt{x}) = 64\]
dan815
  • dan815
|dw:1371328428597:dw|
Loser66
  • Loser66
@dan815 then?
dan815
  • dan815
subsitution
Jhannybean
  • Jhannybean
Oh...I see.
Loser66
  • Loser66
I need more, dan, how to sub?
Jhannybean
  • Jhannybean
Wouldnt that be your answer? :\ Unless you're just going to have \[\large 2x +2\sqrt{x^2+16x}=48\] divide everything by 2. \[\large x+ \sqrt{x^2+16x} = 24\]
dan815
  • dan815
|dw:1371328802389:dw|
dan815
  • dan815
square both sides and solve
Loser66
  • Loser66
got you, dan.
anonymous
  • anonymous
Nope:$$\sqrt{x^2+15x}=24-x\\x^2+15x=x^2-48x+576\\63x=576\\x=\frac{576}{63}=\frac{64}7$$
Jhannybean
  • Jhannybean
You have to complete the square -_-
anonymous
  • anonymous
Now you plug that in to see if it all works.
dan815
  • dan815
u messed up somewhere
anonymous
  • anonymous
I just trusted @Jhannybean 's work from the previous post was correct
dan815
  • dan815
|dw:1371329088136:dw|
dan815
  • dan815
ah ok
dan815
  • dan815
just 1 tiny mistake that 15 gotta be 16 and ull get 9
Jhannybean
  • Jhannybean
Green is not in your favor dan, lol.
dan815
  • dan815
omg big mistake
dan815
  • dan815
|dw:1371329430958:dw|
Jhannybean
  • Jhannybean
\[\large 2x +2\sqrt{x^2+16x}=48\]\[\large 2\sqrt{x^2+16x}=48-2x\]\[\large \sqrt{x^2+16x}=24-x\]\[\large (\sqrt{x^2+16x})^2=(24-x)^2\]\[\large x^2-16 = 576 -48x +x^2 \]\[\large 48x = 576 +16\]\[\large x= \frac{592}{48}=\frac{37}{3}\]
dan815
  • dan815
|dw:1371329446191:dw|
Jhannybean
  • Jhannybean
Wait what...
anonymous
  • anonymous
Yes 9 is the answer and it is not extraneous. It is resolved in this similar manner
1 Attachment
Jhannybean
  • Jhannybean
Damn :|
dan815
  • dan815
ur mistake is 5th line ghanny thats 16x not 16
dan815
  • dan815
and +16x not -
Jhannybean
  • Jhannybean
T_T ....
Jhannybean
  • Jhannybean
I'm going to redo this...
anonymous
  • anonymous
thank you so much for all yoru hard work everyone :)
dan815
  • dan815
JORHE :)
dan815
  • dan815
Jorge*
dan815
  • dan815
do you say it Horhey?
anonymous
  • anonymous
hahaha you say it whoreje
dan815
  • dan815
coooooooooooooooooooOOOOOooooooooooooooooooool :)
dan815
  • dan815
lets be friends!
dan815
  • dan815
i always wanted to know a JORJEE
anonymous
  • anonymous
hahaha sounds great. Im doing homework my friend we shall talk later :) Thanks once again.
dan815
  • dan815
ok tell me if u do some fun problem
dan815
  • dan815
write it here
whpalmer4
  • whpalmer4
@oglon3r the extraneous solution bit: there's not one here, but sometimes when you solve an equation by squaring both sides, you'll get a solution that when substituted back in the original equation doesn't actually solve it. So, whenever you get a solution and you've squared both sides, you need to check that it works. Personally, I think whenever you get a solution to anything you need to check that it works, but especially when you square things...
dan815
  • dan815
WHpalmer is ur name william
dan815
  • dan815
william harrison palmer
Jhannybean
  • Jhannybean
\[\large \sqrt{x+16} + \sqrt{x} = 8\]\[(\large \sqrt{x+16} + \sqrt{x})^2 = (8)^2\]\[\large (x+16) +2\cdot \sqrt{x} \cdot \sqrt{x+16} +x = 64\]\[\large 2x+16 +2\sqrt{x(x+16)}=64\]\[\large 2x+16 + 2\sqrt{x(x+16)}=64\]\[\large 2\sqrt{x(x+16)} = -2x -16+64\]\[\large (2\sqrt{x(x+16})^2 = (-2x +48)^2\]\[\large 4x(x+16)= 4x^2 -192x+2304\]\[\large 4x^2 + 64x=4x^2 -192x +2304\]\[\large 4x^2-4x^2 +64x +192x = 2304 \]\[\large 256x =2304 \]\[\large x=9\]
dan815
  • dan815
yes but loll why wud u actually multiply those numbers out haha
Jhannybean
  • Jhannybean
...YAY!!!! lol.
Jhannybean
  • Jhannybean
Because i take the long method. -_-
dan815
  • dan815
haha
whpalmer4
  • whpalmer4
Here's an example: \[x = \sqrt{2-x}\]Square both sides\[x^2=2-x\]\[x^2+x-2=0\]Factor\[(x+2)(x-1)=0\]Solve for \(x\)\[x=-2, x=1\]Now if we try both solutions in the original equation: \[-2=\sqrt{2-(-2)} \rightarrow-2=2\]\[1=\sqrt{2-1}\rightarrow 1 = 1\]The second solution works, but the first does not. The first solution is extraneous.
whpalmer4
  • whpalmer4
@dan815 my friends call me Bill, or "hey you!" :-)
Jhannybean
  • Jhannybean
I need to learn shortcuts -_-
dan815
  • dan815
yes
dan815
  • dan815
what kind of shortcuts?
whpalmer4
  • whpalmer4
A graphical shortcut for this one is to observe that \(\sqrt{x+16}\) is fairly linear, and can be approximated by \(y = x/10+4\). Then just sketch \(8-\sqrt{x}\) for a few points and you get an excellent idea of where they cross. When you see that, it's immediately obvious that 9 works...
Loser66
  • Loser66
@whpalmer4 the way you solve is really excellent. Question here: \[\sqrt{x+16}\] how to turn to y = s/10 + 4
whpalmer4
  • whpalmer4
Oh, I just observed that at x = 0, y = 4, and at x = 10, y is about 5.

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