(√6+√5)(√6-√5)
2006-07-25 09:52:32 · 4 answers · asked by Brandon ツ 3 in Science & Mathematics ➔ Mathematics
6 - 5 = 1
2006-07-25 09:57:43 · answer #1 · answered by MsMath 7 · 2⤊ 3⤋
Remember factoring? When you had a difference of two squares, it always factored into the conjugates of the roots? Your problem is the same, just in reverse.
(â6+â5)(â6-â5) = (â6)² - (â5)² = 6 - 5 = 1
2006-07-25 17:41:46 · answer #2 · answered by Louise 5 · 0⤊ 0⤋
(â6+â5)(â6-â5)
FOIL:
First: â6 * â6 = 6
Outside: â6 * -â5 = -â30
Inside: +â5 * â6 = +â30
Last: â5 * -â5 = -5
Add them all up:
6 - â30 + â30 - 5
=6 - 5
=1
2006-07-25 19:53:25 · answer #3 · answered by Anonymous · 0⤊ 0⤋
sqrt= square root
sqrt(6)(sqrt(6))-sqrt(5)sqrt(5)
(sqrt6)^2-(sqrt5)^2
6-5=1
the product of the conjugates is 1
steps: since your products are in the form of (x+n)(x-n), you only need to foil out the first and last terms. so you have the sqrt6 squared and the sqrt5 squared. A square root squared is equal to what is inside, so you are left with the 6 and the 5, 6-5=1.
2006-07-25 17:06:40 · answer #4 · answered by pilotmanitalia 5 · 0⤊ 0⤋