2007-07-27 08:32:30 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(sqrt3 - sqrt5)(sqrt3 - sqrt 5)
Use FOIL
= (sqrt3)(sqrt3) + (sqrt3)(-sqrt5) + (-sqrt5)(sqrt3) + (-sqrt5)(-sqrt5)
= 3 - sqrt(15) - sqrt(15) + 5
= 8 - 2sqrt(15)
2007-07-27 08:36:29 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
sqrt3*sqrt3-sqrt3*sqrt5-sqrt5*sqrt3+sqrt5*sqrt5
the first and last terms are product of square root so they equal 3 and 5. The middle 2 terms are the same thing so I'll rewrite them as twice the square root of the product of the radicands
3+5-2(sqrt15)
8-2(sqrt15)
2007-07-27 08:43:26 · answer #2 · answered by MLBfreek35 5 · 0⤊ 0⤋
(√3 - √5)(√3 - √5)
= (√5 - √3)(√5 - √3)
= (√5 - √3)^2
= 5 - 2√15 + 3
= 8 - 2√15
2007-07-27 08:38:57 · answer #3 · answered by psbhowmick 6 · 0⤊ 0⤋
(√3 - √5)(√3 - √5)
=(√3-√5)^2
3-2*√3*√5+5
= 8 - 2√15
2007-07-27 09:19:02 · answer #4 · answered by sss 2 · 0⤊ 0⤋
(√3 - √5) (√3 - √5)
3 - √3 √5 - √3√5 + 5
8 - 2√3 √5
8 - 2√15
2007-07-27 11:32:49 · answer #5 · answered by Como 7 · 0⤊ 0⤋
(sq rt3-sq rt 5)(sq rt3-sq rt5)
(sq rt3-sq rt5)^2
or (sq rt3)^2+(sq rt5)^2-2(sq rt3)(sq rt5)
or 3+5-2(sq rt 15)
or 8-2(sq rt15) ans
2007-07-27 08:43:25 · answer #6 · answered by MAHAANIM07 4 · 0⤊ 0⤋