Verified answer

Is -√64 greater than √64?

No.

All positive numbers have two square roots a positive root and its mirror negative, the square root of 64 has both 8 and negative 8 as answers.

typically a square root sign with a plus minus sign in front (±) means that both roots (positive and negative are acceptable) e.g ±√64

=8

or = -8

usually the positive root of a number is shown by leaving out the plus/minus sign

e.g. √64 (positive root)

= 8

while the negative root is denoted by a minus sign in front of the square root sign

e.g. -√64

= -8

as 8 is not equal to -8

the √64 is not equal to the -√64

No, root 64 is greater than negative root 64. NOTE: negative root 64 and root negative 64 are different. root NEGATIVE 64 is imaginary whilst NEGATIVE ROOT 64 is real, but is not an element of the set of positive whole numbers.

no negative root 64 will give u -8 wheras positive root 64 gives u 8.

positive is always greater than negative so negative 64 is not greater than positive root 64.

They are same because if we take -8^2 it will be -8*-8=+64 or +8^2=+64 -root64=+8=+root64.

-√64 is -8 and √64 is 8. So √64 is bigger than -√64.

There equal because when you square root 64 the answer is just 8 and eight equals eight

It's equal to, because your doing pythagoras therom. 8x8=64 so your answer for √64=8. Therefore it is equal.

√64 = 8

If you put a negative sign in front if it, it becomes -8. -8 is less than 8.

they're both the same. But I can't remember why. :D