How do I square 1 - the square root of 3 halves?

Question

I'm confused. If I use the (a-b) squared - my answer should be a^2 - 2ab+b^2..... But - I must be doing something wrong because I don't have the right answer.... Please help!

Answer 1

(1 - √(3/2))^2 =
1 - 2√(3/2) + (3/2)
(5/2) - 2√(3/2)

Answer 2

I believe the focus of your question is on the +/- 2. This simply means that both +2 and -2 when squared, give you +4. And of course, the imaginary 'i' is there because you cannot take the square root of a negative number. If you were to plot the equation y = x^2 on graphic paper, you would get a parabola and 2 of the points on the parabola would be (2, 4) and (-2, 4). Any equation of the second degree (i.e. the highest exponent is 2) produces a parabola which confirms that for each value of y, there are 2 values of x; Except for the maximum or minimum point which is usually found by taking the derivative of the equation.

Answer 3

(a - b)^2 =
(a - b)(a - b) =
a^2 + b^2 - 2ab - as given.

(1 - sqrt3/2) ^2
(1 - sqrt3/2)(1 - sqrt3/2)
1 + 3/2 - 2sqrt(3/2)
5/2 - 2sqrt(3/2)

NB 3/2 = (sqrt(3/2)^2 = sqrt(3/2) x sqrt(3/2).
When a 'square root' is multiplied to itself it just removes the square root sign, leaving the number under the square root the same.

Answer 4

(1-sqrt(3)/2)
= 1^2 - 2 * 1 * sqrt(3)/2 + (sqrt(3)/2)^2
= 1 - sqrt(3) + 3/4
= 7/4 - sqrt(3)

Answer 5

( 1 - sqrt(3/2) ) ^ 2=

1 - 2 * 1 * sqrt(3/2) + (sqrt(3/2)) ^ 2 =

1 - 2 * sqrt(3/2) + 3/2

Answer 6

(a-b)^2=a^2 - 2ab+b^2 that's right