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Square root of 1/2 = (1)/(square root of 2) = 1/1.4142 = 0.7071 Also Square root of 1/2 = Square root of 0.5 = 0.7071
square root of (x2 + 1) = no simplification (square root of x2) + 1 = x + 1
-27 does not have a square root. The mathematical expression for the square root of-1 is i.
x + y = 1xy = 1y = 1 - xx(1 - x) = 1x - x^2 = 1-x^2 + x - 1 = 0 or multiplying all terms by -1;(-x^2)(-1) + (x)(-1) - (1)(-1) = 0x^2 - x + 1 = 0The roots are complex numbers. Use the quadratic formula and find them:a = 1, b = -1, and c = 1x = [-b + square root of (b^2 - (4)(a)(c)]/2a orx = [-b - square root of (b^2 - (4)(a)(c)]/2aSox = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)x = [1 + square root of (1 - 4]/2x = [1 + square root of (- 3)]/2 orx = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;x = [1 + square root of (i^2 )(3)]/2x = [1 + (square root of 3)i]/2x = 1/2 + [i(square root of 3]/2 andx = 1/2 - [i(square root of 3)]/2Since we have two values for x, we will find also two values for yy = 1 - xy = 1 - [1/2 + (i(square root of 3))/2]y = 1 - 1/2 - [i(square root of 3)]/2y = 1/2 - [i(square root of 3)]/2 andy = 1 - [1/2 - (i(square root of 3))/2)]y = 1 - 1/2 + [i(square root of 3))/2]y = 1/2 + [i(square root of 3)]/2Thus, these numbers are:1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/22. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2Let's check this:x + y = 11/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1xy = 1[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:= 1/4 - [(3)(-1)]/4= 1/4 + 3/4= 4/4=1In the same way we check and two other values of x and y.
The answer is 2i. When dealing with negative square roots, the expression i is used to represent the square root of -1.
The answer is 2i. When dealing with negative square roots, the expression i is used to represent the square root of -1.
The answer is 2i. When dealing with negative square roots, the expression i is used to represent the square root of -1.
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By root, I think you mean square root. The square root of 2 is approx. 1.414. The square root of 9 = 3, so this goes exactly at 3 on the number line. Square root 2 will be less than 1/2 way between 1 and 2 on the number line.
The negative square root of a number is expressed as -√n or -n^½. The question probably may refer to the square root of a negative number. In mathematics there is an identity i² = -1 so that i = √-1 The square root of a negative number can be shown in several ways. For Example : √-36 = √( -1 x 36) = √-1 x √36 The square root of 36 is ±6, and the square root of -1 is i. The solution to √-36 is therefore ±6i.
Sometimes. The square root of 2 is, but the square root of 9 is not.
Mathematicians decided that, since the square root of a negative number does not exist, they would use the first letter of "imaginary" to represent this "value".
friends,root 10 is according to our pythagores thoream.root 10=square of 3 and square of 1 can be represented on a number line.
Negative square root is -√Square root of negative one is i.
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I cant use the square root symbol so lets use the word "root" to represent square root. The answer would be 4root2