Question

If x+y=12 and xy=32, find the value of x^2+y^2

Answer 1

Given:-

• x+y=12

• xy=32

To Find:-

• the value of x^2+y^2.

Solution:-

$x+y=12$

Squaring Both Side

$= > {(x+y)}^{2} = {12}^{2}$

We Will 1st Identity

i.e,

(a+b)²= a²+b²+2ab

$= > {x}^{2} + {y}^{2} + 2xy = 144$

In the Question It is given that xy=32

then, 2xy=64

$= > {x}^{2} + {y}^{2} + 64 = 144$

$= > {x}^{2} + {y}^{2} = 144 - 64$

$= > {x}^{2} + {y}^{2} = 80$

∴ The Required Value is 80.

Answer 2

xy = 32

=> x= 32

y

x+y=12

=> 32 + y =12

y

=> 32 + y² = 12y

=> y² - 12y + 32 = 0

=> y ( y - 4) - 8 (y-4) = 0

=> (y-8)(y-4) = 0

=> y = 8 or y = 4

If y= 8 , then

xy = 32

=> x = 32/8 = 4

Similarly , If y= 4 , then x = 8.

Now ,

x²+ y² = 8² + 4² = 64+16 = 80.

hope it helps..