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secondary 2 | Maths
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please help w both questions!
Date Posted: 3 years ago
Views: 239
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x⁴ + 324
= (x²)² + 18²
= (x² + 18)² - 2x²(18) (property is a²+b² = (a+b)² - 2ab)
= (x² + 18)² - 36x²
= (x² + 18)² - (6x)²
= (x² + 18 + 6x)(x² + 18 - 6x)
(property is a²-b² = (a+b)(a-b))
= (x² + 2(3)x + 3² + 9)(x² - 2(3)x + 3² + 9)
(completing the square, a²+2ab+b² = (a+b)²)
= ( (x+3)² + 9) ( (x-3)² + 9)
So, the fraction becomes : (13²+9)(7²+9)(25²+9)(19²+9)(37²+9)(31²+9)(49²+9)(43²+9)(61²+9)(55²+9) / (7²+9)(1²+9)(19²+9)(13²+9)(31²+9)(25²+9)(43²+9)(37²+9)(55²+9)(49²+9)
= (61²+9) / (1²+9) (after cancelling out common terms)
= 3730/10
= 373
= (x²)² + 18²
= (x² + 18)² - 2x²(18) (property is a²+b² = (a+b)² - 2ab)
= (x² + 18)² - 36x²
= (x² + 18)² - (6x)²
= (x² + 18 + 6x)(x² + 18 - 6x)
(property is a²-b² = (a+b)(a-b))
= (x² + 2(3)x + 3² + 9)(x² - 2(3)x + 3² + 9)
(completing the square, a²+2ab+b² = (a+b)²)
= ( (x+3)² + 9) ( (x-3)² + 9)
So, the fraction becomes : (13²+9)(7²+9)(25²+9)(19²+9)(37²+9)(31²+9)(49²+9)(43²+9)(61²+9)(55²+9) / (7²+9)(1²+9)(19²+9)(13²+9)(31²+9)(25²+9)(43²+9)(37²+9)(55²+9)(49²+9)
= (61²+9) / (1²+9) (after cancelling out common terms)
= 3730/10
= 373
(xy - 3x + 7y - 21)ⁿ
= (x(y - 3) + 7(y - 3))ⁿ
= ((x + 7)(y - 3))ⁿ
= (x + 7)ⁿ (y - 3)ⁿ
Every term in each expansion is unique in the sense that they do not have the same power of x or y. So their products will be unique as well and there is no like term to collect at all.
For any binomial (a + b)ⁿ, there are always n+1 terms.
So for the above 2 expansions, we have n+1 terms for each.
Total number of terms after multiplication
= (n+1)²
( n+1 terms multiplied by n+1 terms )
So we just need to solve (n+1)² ≥ 1996
n is definitely positive so we can just square root both sides.
n+1 ≥ √1996
n+1 ≥ 44.7 (3s.f)
n ≥ 43.7
Since n is a positive integer, n ≥ 44
Smallest n is 44
(Edited n ≥ 45 to n ≥ 44)
= (x(y - 3) + 7(y - 3))ⁿ
= ((x + 7)(y - 3))ⁿ
= (x + 7)ⁿ (y - 3)ⁿ
Every term in each expansion is unique in the sense that they do not have the same power of x or y. So their products will be unique as well and there is no like term to collect at all.
For any binomial (a + b)ⁿ, there are always n+1 terms.
So for the above 2 expansions, we have n+1 terms for each.
Total number of terms after multiplication
= (n+1)²
( n+1 terms multiplied by n+1 terms )
So we just need to solve (n+1)² ≥ 1996
n is definitely positive so we can just square root both sides.
n+1 ≥ √1996
n+1 ≥ 44.7 (3s.f)
n ≥ 43.7
Since n is a positive integer, n ≥ 44
Smallest n is 44
(Edited n ≥ 45 to n ≥ 44)
Amazing solution J :-)
Thanks Celio!
Nancy, please take note of the edit above
Nancy, please take note of the edit above
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