The square root of 0 is 0, which is a real number.
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"0" And if you think about it, zero times zero is still zero. Zero times anything is zero, so this doesn't have an exception when you do the reverse to find the square root.The square root of a number (n) is a number that when multiplied by itself (squared) results in the original number (n). Therefore, the square root of 0 must be 0 because 0 x 0 = 0.
The square root of 49 is 7 and it is a natural number because natural numbers are from 0-infinity
0
The quadratic equation provides solutions for the generalized equation ax2 + bx + c = 0. The solution(s) are (-b +/- square root ( b2 - 4ac)) / 2a. Plugging in a=1, b=3, and c=-15 (from x2 + 3x -15 = 0) you get... (-3 +/- square root (32 - 4(1)(-15)) / 2(1) ... or ... (-3 +/- square root (9 + 60)) / 2 ... or ... (-3 +/- square root (69)) / 2 ... or ... about -3 +/- 8.3 Since 69 is positive, the square root (69) is real, at 8.3 This equation has two real roots, x=5.3 and x=-11.3 If the discriminant (b2 - 4ac) were zero, there would be one real root. If it were negative, there would be one real root and one imaginary root, i.e. a complex conjugate.
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of