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Q: 2 x - 1 2 square?

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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)

The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .

X^2 + X = 0 halve the linear term (1) and square it then add to both sides X^2 + X + 1/4 = 1/4 factor left; gather terms right (X + 1/2)^2 = 1/4 (X + 1/2)^2 - 1/4 = 0 (-1/2,-1/4) vector and the number 1/4 was added to both sides completing the square

(5w4y5)1/2 x (20w6y4)1/2= (5w4y5 x 20w6y4)1/2= (100w10y9)1/2= 10w5y9/2

-1/2*x-3/2 which is equal to -1/[2*x3/2]

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(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

(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

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)

3

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:

To find the square feet, multiply 1 foot by 2 feet. 1 x 2 = 2 square feet.

: x squared times square root of x : : = x^2 * x^(1/2) : = x^[2+(1/2)] : = x^[(4/2)+(1/2)] : = x^(5/2) : which is the same thing as : square root of (x raised to the 5th power)

(sin(x))^2+(cos(x))^2=1

sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))

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.

Whenever you take the square root of a negative number, you get an imaginary number in addition to the other terms.eg:√-1 = i√-2 = =√-(1)(2) =i√2√-100 = 10iSo the square root of -x is:√-x =√(-1)(x) = i√x

A square can't be 1 x 2. In a 1 x 2 rectangle, it's likely that 1 is the width.

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