It might seems like it, but actually no.
Proof:
sqrt(0) = 0 (0 is an integer, not a irrational number)
sqrt(1) = 1 (1 is an integer, not irrational)
sqrt(2) = irrational
sqrt(3) = irrational
sqrt(4) = 2 (integer)
As you can see, there are more than 1 square root of a positive integer that yields an integer, not a irrational.
While most of the sqrts give Irrational Numbers as answers, perfect squares will always give you an integer result.
Note: 0 is not a positive integer. 0 is neither positive nor negative.
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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.
You can find the square root of an irrational number by approximating irrational square roots of them, after you use the calculator. (The calculator gives an approximate root also) For example,1. Approximate the square root of 4.3 to the nearest hundredth.Use the calculator, which shows 2. 0736444135.Since 3 < 5 round down to 2.07 and drop the digits to the right of 7.2. Approximate the negative square root of 10.8 to the nearest hundredth.Use the calculator, which shows -3.286335345Since 6 > 5 round up to -3.29 and drop the digits to the right of 8.
Octagons don't have square roots.
No answer in integers. Quadratic formula gives roots of x2 + x - 26 = 0 as 5.625 and -4.625
To solve the equation x^2 = n, you would take the square root of both sides to isolate x. This would give you x = ±√n, where the ± symbol indicates that there are two possible solutions, one positive and one negative. Therefore, the best way to solve this equation is to take the square root of n and include both the positive and negative roots as solutions.