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What is the derivative of y equals square root of x when x is positive?
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)
X2-6x plus 1 equals 0?
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:
What is x2 plus x-5 equals 0?
x=1/2(-1-square root of 21) and x=1/2(square root of 21 -1)
How do you differentiate square root x?
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
What number do you add to the polynomial x2 plus x to complete the square?
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
How do you factor x squared plus x minus one half?
(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
How do you factor x square plus one?
(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
How many x are there for square root of 1-x equals square root of x-2?
3
X2-6x plus 1 equals 0?
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:
How many square feet is 1' x 2'?
To find the square feet, multiply 1 foot by 2 feet. 1 x 2 = 2 square feet.
Find the two numbers whose product is 1 and whose sum is 1?
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.
What is the derivative of square root of x?
sqrt(x) = x^(1/2) The derivative is (1 / 2) * x^(-1 / 2) = 1 / (2 * x^(1 / 2)) = 1 / (2 * sqrt(x))
Factor the polynomial x2-x-35?
Every polynomial defines a function, often called P. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. So, P(x) = x^2 - x - 35 0 = X^2 - x - 35 or, x^2 - x - 35 = 0 We can factor this equation as (x - r1)(x - r2) = 0. Let's find r1 and r2: x^2 - x - 35 = 0 add 35 to both sides; x^2 - x = 35 ad to both sides 1/4 in order to complete the square; x^2 - x + 1/4 = 35 + 1/4 (x - 1/2)2 = 141/4 x - 1/2 = +,- square root of 141/4 x = 1/2 +,- 1/2(square root of 141) x = (1 + square root of 141)/2 or x = (1 - square root of 141)/2 So the factorization is: [x - (1 + square root of 141)/2 ] [x - (1 - square root of 141)/2 ] Check.
If a square is 1X2 what is the width?
A square can't be 1 x 2. In a 1 x 2 rectangle, it's likely that 1 is the width.
What is the integral of the square root of x minus 1 squared?
the integral of the square-root of (x-1)2 = x2/2 - x + C
How would you find the derivative of the square root of x?
square root of x is the same as x^(1/2). You do the same as you would normally 1/2 * x^(-1/2). Take the exponent, put it down in front of x and subtract the exponent by 1 (1/2 - 1 = -1/2)
How would you work out 9 to the power of 1 slash 2 i know it makes 3 but how?
One of the laws of indices states thatx^a * x^b = x^(a+b)Now if you put a = b = 1/2, you getx^(1/2) * x^(1/2) = x^(1/2+1/2) = x^1 = xSo x^(1/2) multiplied by itself = x and therefore x^(1/2) is the square root of x.Then, since the square roots of 9 are 3 and -3 you have the answer to your question.
