Question

${x}^{2} + {y}^{2} = 47 and xy = \frac{19}{2} , then the value of 2 (x + y) ^{2} is _____$
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Answer 1

Required Answer:-

Given:

• x² + y² = 47
• xy = 19/2

To find:

• The value of 2(x + y)²

Answer:

• 2(x + y)² = 132

Solution:

Given that,

➡ x² + y² = 47 ......(i)

➡ xy = 19/2

➡2xy = 19 ...........(ii)

Adding equations (i) and (ii), we get,

➡ x² + 2xy + y² = 47 + 19

➡ (x + y)² = 66

➡ 2(x + y)² = 2 × 66

➡ 2(x + y)² = 132

Hence, the value of 2(x + y)² is 132.

Answer:

• 2(x + y)² = 132

Identity Used:

• (x + y)² = x² + 2xy + y²

Other Identities:

• (x - y)² = x² - 2xy + y²
• x² - y² = (x + y)(x - y)
• (x + y)³ = x³ + 3x²y + 3xy² + y³
• (x - y)³ = x³ - y³ - 3x²y + 3xy²

Answer 2

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