Question

if x+y=4 and √xy=2 then find x²+y²​

Answer 1

Answer:

√xy=2→xy=4→2xy=8 (1)

x+y=4→(x+y)^2=16→x^2+y^2+2xy=16 (2)

We place relationship one in relation to two

x^2+y^2+8=16→x^2+y^2=8

Step-by-step explanation:

Answer 2

Answer:

$x^{2} +y^{2}$ = 8

Step-by-step explanation:

given:

$x^{2} +y^{2}$ = ?

$x+y$ = 4

$\sqrt{xy}$ = 2     ⇒  $xy$  = 4

we know that ,

$( x + y)^{2} = x^{2} + 2xy + y^{2}$

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

           =  $(4^{2})- 2(4)$

          = 16 - 8

          = 8

therefore , $x^{2} +y^{2}$ = 8

Hope it helps !