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Q: Square root of 0

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x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!

Any number greater than 0 has two square roots, a positive square root and a corresponding negative square root. Rounded to two decimal places, the square roots of 200 are ±14.14.

the derivative is 0. the derivative of a constant is always 0.

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.

if this is x squared -6x+6=0 then -6=b, a=1, c=6 6+ square root of -6 squared-4(6x1) - 6+ square root of (36-24) - 6+ square root of 12 - 6+ square root of 4 x square root of 3 - 6 + (2x square root of 3) - that is all divided by 2 multiplied by a meaning it is divided by 2. so x= 6 + or - (2 square root 3) divided by 2 srry steps are jmbled -

Related questions

The square root of 1 is 1.The square root of 0 is 0.

Sqrt(0) = 0 so the answer is 0.

The square root of 0 is 0, which is a real number.

If x is 0, the square root is 0 also.

The square root of 0 is 0. Since 0 has no positive or negative equivalent, this is its only square root.

0

The answer is 0.

The square root of both 0 and 1 equals the square of 0 and 1

sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of

0

0 = √0.

0 0 x 0 = 0

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