Simple algebraic manipulation

We use $(x+y)^{2} = x^{2} + y^{2} + 2xy$ after substituting the given values in the equation we get the answer as 1 which is correct.

x+y=3
(x+y)^2=9
x^2+y^2=9-2xy
xy=4
so x^2 +y^2=9-8=1

$x^2 +y^2 =(x+y)^2 -2xy=3^2-4 \times 2=1$

Given that,

$x + y = 3$

$xy = 4$

Now,

$x^2 + y^2 = ?$

$=(x + y)^2 - 2xy$

$= (3)^2 - 2 \times 4$

$\Rightarrow 9 - 8 = 1$

x^2 + y^2 = (x^2 + 2xy + y^2) - 2xy = (x + y)^2 - 2xy = 3^2 - 2(4) = 9 - 8 = 1

First we should square on both sides then we get( x+y)"2 =9 X"2 +y"2 + 2xy =9 Substituting xy =4 X"2 +y"2 +2×4=9 X"2+y"2=9-8 X"2+y"2=1

x + y = 3 ( x + y )² = 9 x² + y² + 2xy = 9 x² + y² = 9 - 2xy x² + y² = 9 - 8 x² + y² = 1

(x + y)2 = x2 + y2 + 2xy; (3)2 = x2 + y2 + 2x4; 9 = x2 + y2 + 8; 9 - 8 = x2 + y2; 1 = x2 + y2