(5+sqrt5)^2+(5-sqrt5)^2. Solve.
2007-05-25 16:54:48 · 7 answers · asked by MathG 1 in Science & Mathematics ➔ Mathematics
(5+sqrt5)^2+(5-sqrt5)^2
we need to FOIL each square
(5+sqrt5)^2 = (5+sqrt5)(5+sqrt5)^2 = 25+5sqrt5+5sqrt5+5
remember sqrt5*sqrt5 = 5
(5-sqrt5)^2 = (5-sqrt5)(5-sqrt5)^2 = 25-5sqrt5-5sqrt5+5
(5+sqrt5)^2+(5-sqrt5)^2
25+10sqrt5+5 + 25 - 10 sqrt5 + 5
60
=]
..
2007-05-25 16:58:08 · answer #1 · answered by Anonymous · 1⤊ 1⤋
We know that,
(a+b)^2+(a-b)^2=2(a^2+b^2)
[if we now accept 5 as a and sqrt5 as b,we can solve the problem easily ]
Therefore.
(5+sqrt5)^2+(5-sqrt5)^2
=2{(5)^2+(sqrt5)^2}
=2(25+5)
=2*30=60
2007-05-25 17:41:29 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
(5 + â5)² = 25 + 10.â5 + 5 = 30 + 10.â5
(5 - â5)² = 25 - 10.â5 + 5 = 30 - 10.â5
Sum = 60
2007-05-26 09:46:18 · answer #3 · answered by Como 7 · 0⤊ 0⤋
60
2007-05-25 17:16:49 · answer #4 · answered by astig 1 · 0⤊ 0⤋
60
2007-05-25 16:58:51 · answer #5 · answered by The Franchise 1 · 0⤊ 0⤋
(5+sqrt5)^2=52.3
(5- sqrt5)^2=7.6
52.3+7.6=59.9
2007-05-25 19:47:19 · answer #6 · answered by Mehdi 3 · 1⤊ 0⤋
25+s^2q^2r^2t^2 25+25+s^2q^2r^2t^2 25
50+2s^2 2q^2 2r^2 2t^2 50
the (-) on the second set is positive because of the exp of 2 then you just add the two binomials together
2007-05-25 17:02:57 · answer #7 · answered by Kristenite’s Back! 7 · 0⤊ 1⤋