Question

if x+y=12 and xy =32 then find the value of x²+ y²​

Answer 1

Answer:

ok

x=12-y

XY=32

(12-y)y=32

12y-y^2=32

y^2-12y+32=0

y^2-8y-4y+32=0

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

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

y=4 or 8

then

x=8 if y=4

then

x^2+y^2=8*8+4*4=64+16=80

is the answer

Answer 2

Answer:

given =.x+y = 12

xy=32

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

x+y = 12 (squaring on both side)

$x^{2}+y^{2}+ 2xy = 144 \\ x^{2}+y^{2}+2×32 = 144 \\ x^{2}+y^{2}=.144-2×32 \\ x^{2}+y^{2}=80$