Question

if alpha+beta=2, alpha×beta=1, then(alpha-beta) ^2 is​

Answer 1

Let x be any positive integer and y = 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = m

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ………. (2)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 ………. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,x2= 3m + 1……… (3)

… (3)Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 2

Answer:

0

Step-by-step explanation:

(alpha-beta)^2           (a-b)^2= a^2+b^2-2ab

= alpha^2 + beta^2 - 2(alpha)(beta)

here alpha^2+beta^2 can be written as (alpha+beta)^2 - 2(alpha)(beta) (this is actuall a formula so try to remember it as it can help u in the future)

= (alpha+beta)^2 - 2(alpha)(beta) - 2(alpha)(beta)

= (alpha+beta)^2 - 4(alpha)(beta)

= 2^2 -4(1)

=4 - 4 =0

Hope this helps!!