f(x)=x2
f -1(x)=sq. root(x)
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y2=x [invert y]
(y2)-1=x = 1/y2=x
sq. root(1/y2)= sq. root (x) [y can't be squared, root both sides]
(1/y)=sq. root(x)
y-1=sq. root(x)
The function f(x)=100 has no roots, as f(x) never equals 0. The square root of 100 is 10. The cube root of 100 is about 4.64.
There is no formula relating to a perfect square but if you want a method 1. Find square root(x) 2. Take the integer component (integral value) of square root(x) 3 Add 1 to intenger(square root(x)) 4. square it So: (integer(square root(x)) + 1)^2
the integral of the square-root of (x-1)2 = x2/2 - x + C
Finding the square root of a number is the same as saying: What number multiplied by itself is equal to the number. In this instance, the square root of 1 is equal to 1 x 1. Therefore, the square root of 1 is 1.Similarly, the cube root of 1 is also 1 (1 x 1 x 1 = 1)
If you have a non-scientific calculator you can use the Newton-Raphson method. Suppose you wish to find the square root of 7. Let f(x) = x2 - 7 so that f(x) = 0 when x is the square root. That is, you want to find x such that f(x) = 0. Let f'(x) = 2x [f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.] Make a guess at the square root of 7, and call is x0. Then calculate xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ... Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. It does so even if your first guess is not so good. Suppose you start with x0 = 5 (a pretty poor choice since 52 is 25, which is nowhere near 7). Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.
1/(2*(square root of x))
Yes, because when x equals 1, the square root of x is rational and the square root of -x is irrational, and when x equals -1, the square root of x is irrational and the square root of -x is rational.
square root of (x2 + 1) = no simplification (square root of x2) + 1 = x + 1
Yes.
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
Yes, if x and y = 1 √1 + √1 = √1 + 1 1 + 1 = 1 + 1 QED
The square root of x can also be written as x^1/2. The cubic root of x is x^1/3, the fourth root x ^1/4, and so on.
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
The Newton-Raphson method is a pretty efficient process. See link for details. If you want the square root of 34, say, define f(x) = x2 - 34 so that when x is the square root of 34, f(x) is zero. That would imply f'(x) = 2x So start with x0 and use the iteration xn+1 = xn - f(xn)/f'(xn) = xn - [xn2 - 34]/[2*xn]
X^1/2 + Y^1/2 Essentially it is just the square root of x plus the square root of y, there's no real way of simplifying that.
x2 - 6x + 1 = 0 x = [-(-6) +&- square root of [(-6)2 - 4(1)(1)]]/2(1) x = (6 +&- square root of 32)/2 x = [6 +&- 4(square root of 2)]/2 x = 3 +&- 2(square root of 2) x = 3 + 2(square root of 2) or x = 3 - 2(square root of 2) Check:
y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)