If 4x² + y² = 40 and cy = 6 , find the value of 2x + y.
Answers
Solution -
If 4x² + y² = 40 and xy = 6, find the value of 2x+y .
4x² + y² = 40
4x² + y² + 4xy = 40 + 4xy
> (2x)² + 2(2x)y + y² = 40 + 4xy
> (2x+y)² = 40+4xy
> (2x+y)² = 40+24 = 64
> (2x+y) = ±8 .
This is the required answer.
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Additional Information -
For integers a, b, c € N, the following identities hold true -
(a + b)² = a² + 2ab + b²
(a + b)² = (a - b)² + 4ab
(a - b)² = a² - 2ab + b²
(a - b)² = (a + b)² - 4ab
a² + b² = (a + b)² - 2ab
a² + b² = (a - b)² + 2ab
2 (a² + b²) = (a + b)² + (a - b)²
4ab = (a + b)² - (a - b)²
ab = {(a + b)/2}² - {(a-b)/2}²
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b)³ = a³ + 3a²b + 3ab² b³
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - 3a²b + 3ab² - b³
a³ + b³ = (a + b)( a² - ab + b² )
a³ + b³ = (a + b)³ - 3ab( a + b)
a³ - b³ = (a - b)( a² + ab + b²)
a³ - b³ = (a - b)³ + 3ab ( a - b )
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Step-by-step explanation:
Solution -
If 4x² + y² = 40 and xy = 6, find the value of 2x+y .
4x² + y² = 40
4x² + y² + 4xy = 40 + 4xy
> (2x)² + 2(2x)y + y² = 40 + 4xy
> (2x+y)² = 40+4xy
> (2x+y)² = 40+24 = 64
> (2x+y) = ±8 .
This is the required answer.
________________________________________
Additional Information -
For integers a, b, c € N, the following identities hold true -
(a + b)² = a² + 2ab + b²
(a + b)² = (a - b)² + 4ab
(a - b)² = a² - 2ab + b²
(a - b)² = (a + b)² - 4ab
a² + b² = (a + b)² - 2ab
a² + b² = (a - b)² + 2ab
2 (a² + b²) = (a + b)² + (a - b)²
4ab = (a + b)² - (a - b)²
ab = {(a + b)/2}² - {(a-b)/2}²
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b)³ = a³ + 3a²b + 3ab² b³
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - 3a²b + 3ab² - b³
a³ + b³ = (a + b)( a² - ab + b² )
a³ + b³ = (a + b)³ - 3ab( a + b)
a³ - b³ = (a - b)( a² + ab + b²)
a³ - b³ = (a - b)³ + 3ab ( a - b )
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