when x is a negative number --- is a wrong answer since square root of a negative number is not defined. So x has to be zero or a positive number. The correct answer is that when x lies between 0 and 1 (with both limits excluded), its square root is greater than the number itself. Of course at both limits, the square root (assuming the positive square root - since a square root of a number can be positive or negative, both with the same absolute value) is the same as the number.
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.
What is the limit as x approaches infinity of the square root of x? Ans: As x approaches infinity, root x approaches infinity - because rootx increases as x does.
Restricting the discussion to real numbers for now. In this case, the square root is only defined for non-negative numbers; the principal square root of a positive number is the POSITIVE square root. For example, both +5 and -5 are square roots of 25, since both - when squared - give you 25. But the positive square root (+5) is called the principal square root, and if you write the square root symbol, that's the number usually meant.
To simplify x divided by the square root of x, we can rewrite the square root of x as x^(1/2). Dividing x by x^(1/2) is equivalent to multiplying x by x^(-1/2), which simplifies to x^(1-1/2). Therefore, x divided by the square root of x simplifies to x^(1/2), which is the square root of x.
The square root of 6 is an irrational number, approximately equal to 2.44948974278. When you take the square root of 6 and continue to do so to infinity, the number will not converge to a specific value but will approach the square root of 6. This means that as you take the square root of the result repeatedly, the number will get closer and closer to approximately 2.44948974278 but will never exactly reach it.
16
For all the values of x that are less than one and greater than zero.
only if x is greater than 1
√x > 0 when x > 0, that is when x is a positive number.
Sqroot x > x if x<1.
it is greater than and equal to 3
10 is greater than square root 20 √20 = √(2 x 10) = √2 x √10 10 = √10 x √10 √10 > √2 → 10 > √20
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.
say x=-2, the x^2=4, but the square root of 4 is 2 because we always take the positive value, known as the principal root. Using this square root of x^2=|x|. So if x is greater than or equal 0, than square root of x^2 is x, but if x is less than zero we must take its abolute value.
f(x) = (x)^ (1/2) (i.e. the square root of x)
No. The square roots of numbers between 0 and 1 (not including 0) are greater than or equal to (in the case of 1) the number. The square root of 0.49 is 0.7 for example.
x can be any number greater than 16 and less than 81.