Question

If (x-y)^2=100 and xy=20. what's the value of x^2+y^2?

Answer 1

$(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2(xy)\\\\\text{We have}\ (x-y)^2=100\ \text{and}\ xy=20.\\\\\text{Substitute}\\\\100=(x^2+y^2)-2(20)\\100=(x^2+y^2)-40\qquad\text{add 40 to both sides}\\\\140=(x^2+y^2)\\\\Answer:\ \boxed{x^2+y^2=140}$

Answer 2

Answer: The value of $x^2+y^2$ is 140

Step-by-step explanation:

We are given an expression:

$(x-y)^2=100$

And,

xy = 20

Using the identity:

$(a-b)^2=a^2+b^2-2ab$

Solving the expression:

$x^2+y^2-2xy=100$

Putting the value of 'xy' in above equation, we get:

$x^2+y^2-2(20)=100\\\\x^2+y^2=100+40\\\\x^2+y^2=140$

Hence, the value of $x^2+y^2$ is 140