Question

x^2+y^2=47 & xy=19/2 what is the value of this 3(x+y)^2+(x-y)^2

Answer 1

Answer:

207

Step-by-step explanation:

x² + y² = 47 (Given)

xy = 19/2 (Given)

Find (x + y)²:

(x + y)² = x² + y² + 2xy

(x + y)² = 47 + 19/2

(x + y)² = 113/2

Find (x - y)²:

(x - y)² = x² + y² - 2xy

(x - y)² = 47 - 19/2

(x - y)² = 75/2

Find 3(x + y)² + (x - y)²:

3(x + y)² + (x - y)² = 3(113/2) + 75/2

3(x + y)² + (x - y)² = 207

Answer: 207

Answer 2

x² + y² = 47 and xy = 19/2

we know , (x + y)² = x² + y² + 2xy

(x + y)² = 47 + 2 × 19/2 = 47 + 19 = 66

hence, (x + y)² = 66 .......(1)

again, (x - y)² = x² + y² - 2xy

(x - y)² = 47 - 2 × 19/2 = 47 - 19

(x - y)² = 28 .......(2)

now, 3(x + y)² + (x - y)²

from equations (1) and (2),

= 3 × 66 + 28

= 198 + 28

= 226

hence, 3(x + y)² + (x - y)² = 226