What is (sqrt(5)-1)^2?

Question

Would it be 5+1? Or 6? Or would it be like 2sqrt(5)-1? Or what?

I need it without decimals as well, btw.

And I'm studying for a test, this isn't h.w., and I do most of my h.w. I just hate trig functions.

Answer 1

(sqrt(5)-1)^2

(sqrt(5)-1)(sqrt(5)-1)

5-(sqrt[5])-(sqrt[5])+1

5-2(sqrt[5])+1

6-2(sqrt[5])


Answer = 6-2(sqrt[5])

Answer 2

5+1

Answer 3

If tan A = 2, then sin A is plus or minus 2/sqrt(5). That is to say, there's the "first quadrant" case and the "third quadrant" case. First quadrant case: tan A = 2, sin A = 2/sqrt(5), sec A = sqrt(5), cosec A = sqrt(5)/2 (just draw a triangle to see this) Then sinAsecA + tanA - cscA = 2 + 2 - sqrt(5)/2 = 8/2 - sqrt(5)/2 = (8-sqrt(5))/2. I don't agree that it's (12-sqrt(5))/2. Third quadrant case: tan A = 2, sin A = -2/sqrt(5), sec A = -sqrt(5), csc A = -sqrt(5)/2 (again draw a triangle) Then sinAsecA + tan A - cscA = 2 + 2 + sqrt(5)/2 = (8 + sqrt(5))/2. Again I don't agree that it's (12-sqrt(5))/2.

Answer 4

(√5 - 1)² = (√5)² - 2√5 + 1
= 5 - 2√5 + 1
= 6 - 2√5

Answer 5

[ √5 - 1 ]^2
= 5 + 1 - 2√5
= 6 - 2√5.

Answer 6

(sqrt(5) - 1)^2

expand as in the case of (a-b)^2

sqrt(5)^2 - 2sqrt(5)*1 + 1

=>5 - 2sqrt(5) +1

6 - 2sqrt(5)

Answer 7

(√5-1)² = 5+1 -2√5 = 6 - 2√5.
Just square it like any binomial, and use (√5)² = 5.

Answer 8

image it is the expansion of (sqrt 5 -1 )
= (sqrt 5 -1 ) x (sqrt 5 -1)
= sqrt 5 x sqrt 5 -2 sqrt 5 +1
= 6-2 sqrt 5
=3-sqrt 5
its about 0.8 without a calculator

Answer 9

it would be sqrt(5)^2-2sqrt(5)+1 if expanded 5-2sqrt(5)+1

sqrt(5)>=2 but <=3 so make an estimate

Answer 10

you start inside

(sqrt(5) - 1) * (sqrt(5) - 1)
USE THE FOIL Method

sqrt(5)*sqrt(5) + -1*-1 + sqrt(5)*-1 + sqrt(5)-1

5 + 1 - sqrt(5) - sqrt(5)

6 - 2sqrt(5)

Answer 11

(sqrt(5) - 1)^2 = Sqrt(5)^2 - 2(sqrt(5))(1) + 1^2
= 5 - 2(Sqrt(5)) + 1
= 6 - 2(sqrt(5))