root2+root3+root5, root2-root3+root5, root2+root3-root5, -root2+root3+root5.
give the explanation steps also plz.
2007-05-27 02:55:39 · 4 answers · asked by Steve 1 in Science & Mathematics ➔ Mathematics
Using:
(a + b)*(a - b) = a^2 - b^2 you can multiply 1-st and 3-rd terms:
(root(2) + root(3) + root(5))*(root(2) + root(3) - root(5)) =
(root(2) + root(3))^2 - 5 =
2 + 2*root(6) + 3 - 5 =
2*root(6)..........(1)
The second multiplied forth term gives:
(root(5) + (root(2) - root(3)))*(root(5) - (root(2) - root(3))) =
5 - (root(2) - root(3))^2 =
5 - (2 - 2*root(6) + 3) =
2*root(6).........(2)
Multiplying (1) and (2) you get 24 for your final result.
2007-05-27 03:23:29 · answer #1 · answered by fernando_007 6 · 0⤊ 1⤋
Answer : 24
(root2+root3+root5)( root2-root3+root5)( root2+root3-root5)(-root2+root3+root5)=?
Use distributive property and a^2-b^2 expansion
(root2+root3+root5)(oot2+root3-root5)=
2+3+2sqrt(6)-5=2sqrt(6)
(root2-root3+root5)(-root2+root3+root5)
=-2+root6+root10+root6-3-root15-root10+root15+5
=2root6
Hence
2sqrt(6) * 2root6=4*6=24
2007-05-27 03:12:18 · answer #2 · answered by iyiogrenci 6 · 0⤊ 1⤋
8*3*0
0
2007-05-27 03:14:04 · answer #3 · answered by PeHLi KiRaN 2 · 0⤊ 1⤋
the answers is: factor 11y
2007-05-27 03:04:32 · answer #4 · answered by Cloee Quips 4 · 0⤊ 1⤋