Question

if x+y =10 and xy=16 find the value of x^2-xy+y^2 and x^2+xy+y^2

Answer 1

Using, (a + b)² = a² + b² + 2ab,

so,

(x + y)² = (10)²

=> x² + y² + 2xy = 100

=> x² + y² + 2(16) = 100

=> x² + y² + 32 = 100

=> x² + y² = 100 - 32

=> x² + y² = 68

Now simply, we know the value of x² + y² and xy we can find the value of x² + y² + xy and x² + y² - xy

So, x² - xy + y²

=> x² + y² - xy

=> 68 - 16 (since, x² + y² = 68)

=> x² - xy + y² = 52

Now, x² + xy + y²

= x² + y² + xy

= 68 + 16

=> x² + y² + xy = 84


Hope it helps dear friend ☺️✌️

Answer 2

hey there

Using, (a + b)² = a² + b² + 2ab,

(x + y)² = (10)²=>

x² + y² + 2xy = 100 =>

x² + y² + 2(16) = 100=>

x² + y² + 32 = 100

x² + y² = 100 - 32=> x² + y² = 68

( x - y)^2 = x² - xy + y² =

x² + y² - xy = 68 - 16

x² - xy + y² = 52.

x² + xy + y²= x² + y² + xy= 68 + 16 => x² + y² + xy = 84

hope it helps you

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