Find the solution of 4 times the square root of the quantity of x plus 2 equals negative 16, and determine if it is an extraneous solution. x = 14; not extraneous x = 2; extraneous x = 2; not extraneous x = 14; extraneous

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Find the solution of 4 times the square root of the quantity of x plus 2 equals negative 16, and determine if it is an extraneous solution. x = 14; not extraneous x = 2; extraneous x = 2; not extraneous x = 14; extraneous

Mathematics
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\[4\sqrt{x+2}=-16\]
any ideas on how to start?
You would move -16 to the left side, and set it equal to 0?

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I'd divide by the 4.
It's not a polynomial, so you don't have to set it equal to 0
So I'd have? \[\frac{ \sqrt{x+2} }{ 4 }=-4\]
|dw:1441897429536:dw|
OH yeah, the 4 cancel eachother out on the left side,
ok. so the everything left on the left side is under the radical, and the number on the right side is negative, so this equation has no real solution. And the solution must be extraneous
because you can't take a square root and get a negative number as the result
Okay,
next is to square both sides
So x + 2 = 16
right
Then you isolate your variable, to make it x = 14, extraneous.
Thanks :)
you're welcome
What would be the results, if you divided both sides by a -4?|dw:1441898080802:dw|
Squaring both sides getting: x+2 = 16 x=14 \[4\sqrt{14 + 2}=-16\]\[4\sqrt{16}=-16\]\[4\times \pm4=\pm16\]\[\pm16=\pm16\] If we accept that the square root of 16 is a plus or minus 4, then it would seem that 14 is not extraneous Hmmmmm
With that I will log off and do some review.....

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