√x^2+16=5
A. -3
B. 3
C. ±3
D. No solution
I really want to understand how to do this, so can you please explain how you got the answer? I don't understand what to do with the x^2, but I know how to do the rest of the equation. You square both sides, then subtract 16 from both sides. But I don't know what to do with the x^2
Thanks so much for your help
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Verified answer
sqaure both sides
x^2+ 16 = 25
x^2 = 9
x= +/- 3
hence C
the square root of x squared is simply x. Now if that square root symbol goes over the 16 too, then the square root of x^2+16 is x + or - 4, leaving you with x+4=5 and x-4 =5, so x=1 and 9. If the square root only goes over the x^2, then x+16=5, so x=-11.
â(x^2 + 16) = 5
square both sides
x^2 + 16 = 25
subtract 16
x^2 = 9
Take the square root of both sides.
|x| = 3
x = ±3
Check that both solutions are valid. They are. Sometimes squaring something can add extraneous solutions.
The answer is C
you extract the square root both sides: there are 2 solutions -3 and +3...! So C
(sqrt x^2+16=5)^2
x^2+16=25
x^2=-9
D +/- 3i