The square root of 0 is 0, which is a real number.
"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
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of
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.
The square root of a real number is not always positive. The square root of any positive number is positive, the square root of zero is zero (not positive), and the square root of a negative number is complex (i.e. neither positive nor negative). The square root of 16 = -4 or 4. The square root of 0 = 0 The square root of -16 = -4i or 4i
The square root of -1 is not a real number like -2.5, 0, or 5. Instead, it is and imaginary number, i, and i = the square root of -1. The answer is imaginary because you can never take a real number, square it, and get a negative number. However, i^2 = -1.
The square root of 1 is 1.The square root of 0 is 0.
a is a real number since we use just letters to represent real numbers. if a > 0, then its square root is also a real number, so it has two square roots, one positive and one negative. Be careful when you use the radical sign, because it is looking only for the principal square root of a, which is the positive one. if a < 0, then its square root is an imaginary number.
The square root of any negative number is imaginary. No integers nearby.
There is no answer to this problem unless x is 0. For the suare root of 98x to be a real number, x has to be positive or zero. For the square root of -147x to be a real number, x has to be negative or zero. Seeing has x has to fit both requirements, the problem has an answer only if x is zero.
Yes, if the number whose square root we are taking is greater than 0. Only if you try to take the square root of a negative number will you get back an imaginary number. Square roots are often irrational, but that's different from real versus imaginary.
X is greater than 5. If X were equal to 5, X-5 would be 0 which has no real square root. If X were 4, X-5 would be -1 which has no real square root, and so on.
BeginRead:aSet Sqt=pow(a,1/2)If (a
The square root of both 0 and 1 equals the square of 0 and 1
The square of any real number is non-negative. So no real number can have a negative square. Consequently, a negative number cannot have a real square root. If the discriminant is less than zero, the quadratic equation requires the square root of that negative value, which cannot be real and so must be imaginary.