there is a radical sign in front of both 3's (looks like a check mark)
2007-04-06 04:00:15 · 7 answers · asked by carolsilveyderrick1956 1 in Science & Mathematics ➔ Mathematics
[(sqrt 3) + 1] / [(sqrt 3) - 1]
multiply the top and bottom by [(sqrt 3) + 1] which is the conjugate of the denominator. (Same terms, but different sign in the middle.)
[(sqrt 3) + 1][(sqrt 3) + 1] / [(sqrt 3) - 1][(sqrt 3) + 1]
=[(sqrt 3)^2 + 2 sqrt 3 + 1] / [(sqrt 3)^2 - 1]
=(3 + 2 sqrt 3 + 1) / (3 - 1)
=(4 + 2 sqrt 3) / 2
=2(2 + sqrt 3) / 2
=2 + sqrt 3
2007-04-06 04:13:53 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
1 / √3 = √3 / 3
2016-04-01 00:30:29 · answer #2 · answered by ? 4 · 0⤊ 0⤋
[sqrt(3) + 1]/[sqrt(3) - 1]
Note that (a - b)(a + b) = a^2 - b^2
To turn the denominator into a simple number, multiply both numerator and denominator by sqrt(3) + 1
[(sqrt(3) + 1)(sqrt(3) + 1)]/[sqrt(3) - 1)(sqrt(3) + 1)]
[3 + 2sqrt(3) + 1]/[3 - 1]
[4 + 2sqrt(3)]/2
[2(2 + sqrt(3)]/2
2 + sqrt(3)
2007-04-06 04:22:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋
[(sqrt 3)+1]/[(sqrt3)-1
multiply by (sqrt 3) +1
[(sqrt 3)+1]^2/ 3-1
[(sqrt 3)+1]^2/2
2007-04-06 04:15:38 · answer #4 · answered by jaybee 4 · 0⤊ 0⤋
[sqrt(3) + 1]/[sqrt(3) - 1]
Multiply by [sqrt(3) + 1] / [sqrt(3) + 1]
={[sqrt(3) + 1]^2} / 2
2007-04-06 04:14:40 · answer #5 · answered by trojanknight_96 3 · 0⤊ 0⤋
Is this (SQRT(3)+1) / (SQRT(3)-1)
Or SQRT(3+1) / SQRT(3 - 1)
?
(SQRT(3) + 1) / (SQRT(3) - 1) X (SQRT(3) + 1) / (SQRT(3) + 1)
= (SQRT(3) + 1)^2 / ((SQRT(3) - 1)(SQRT(3)+1))
= (SQRT(3)^2 + 2.SQRT(3) + 1) / (3 - 1)
= (3 + 1 + 2SQRT(3)) / 2
= 2(2 + SQRT(3))/2
= 2 + SQRT(3)
2007-04-06 04:14:21 · answer #6 · answered by Orinoco 7 · 0⤊ 0⤋
Square root of 8
Please give me best answer thanks!
2007-04-06 07:00:07 · answer #7 · answered by Anonymous · 0⤊ 0⤋