## anonymous one year ago 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

1. anonymous

$4\sqrt{x+2}=-16$

2. anonymous

any ideas on how to start?

3. anonymous

You would move -16 to the left side, and set it equal to 0?

4. anonymous

I'd divide by the 4.

5. anonymous

It's not a polynomial, so you don't have to set it equal to 0

6. anonymous

So I'd have? $\frac{ \sqrt{x+2} }{ 4 }=-4$

7. anonymous

|dw:1441897429536:dw|

8. anonymous

OH yeah, the 4 cancel eachother out on the left side,

9. anonymous

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

10. anonymous

because you can't take a square root and get a negative number as the result

11. anonymous

Okay,

12. anonymous

next is to square both sides

13. anonymous

So x + 2 = 16

14. anonymous

right

15. anonymous

Then you isolate your variable, to make it x = 14, extraneous.

16. anonymous

Thanks :)

17. anonymous

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