Question

If 5x - 2y = 2 and xy = 2 , find the value of ( 5x - 2y )^2.​

Answer 1

Answer:the value of (5x-2y)^2 is 25x^2+4y^2+36=0

Step-by-step explanation:_if 5x-2y=2

Squaring both sides

(5x-2y)^2=(2)^2

(5x)^2+(-2y)^2-2*(5x)(-2y)=4

25x^2+4y^2+20xy=4

25x^2+4y^2+20(2)=4

25x^2+4y^2+40-4=0

25x^2+4y^2+36=0

Hope it will help you.

Answer 2

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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