if5x-2y=7 and xy=2,find the value of (5x+2y)^2
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Answer:129
Step-by-step explanation:Given:
5x-2y= 7
XY= 2
5x-2y = 7
Squaring on both sides
(5x-2y)² = (7 )²
[( a-b)²= a²+b²-2ab]
(5x)² + (2y)² - 2× 5x× 2y = 49
(5x)² + (2y)² - 20xy= 49
(5x)² + (2y)² - 20 (2) = 49 [XY=20]
(5x)² + (2y)² - 40= 49
(5x)² + (2y)² = 49+ 40= 89
(5x)² + (2y)² = 89..............(1)
We have to find (5x+2y)²
(5x+2y)² = (5x)² + (2y)² + 2× 5x× 2y
[( a+b)²= a²+b²+2ab]
(5x+2y)² = (5x)² + (2y)² + 2× 5x× 2y
(5x+2y)² = (5x)² + (2y)² + 20xy
Put the value of eq 1
(5x+2y)² = 89 + 20 × 2
(5x+2y)² = 89+ 40 = 129
(5x+2y)² =129
Hence, the value of (5x+2y)² = 129
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Hope this will help you.....
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