SOLUTION: Could someone help! If sqrt x^2-16=3 then what is the x must approximately be ?

Question 64415: Could someone help!
If sqrt x^2-16=3 then what is the x must approximately be ?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
I'm not sure if I understand the problem. Is it?:
(1) sqrt(x^2-16)=3
or is it?:
(2) sqrt(x^2)-16=3
We'll solve it both ways. First, to solve for x:
sqrt(x^2-16)=3 We will first get rid of the radical by squaring both sides and we get:
x^2-16=9 add 16 to both sides:
x^2=9+16 =25 take sqrt of both sides
x=+or-5
ck
sqrt(25-16)=3
sqrt(9)=3
3=3
now the other one:
sqrt(x^2)-16=3 first, we'll add 16 to both sides and we get:
sqrt(x^2)=19 now we'll square both sides to get rid of the radical:
x^2=361 take sqrt of both sides
x=sqrt(361)
x=+or-19
ck
sqrt(19^2)-16=3
19-16=3
3=3
Hope this helps. Happy holidays.-----ptaylor

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