Question

If x+y=12 and xy = 32, find the value of
x²+y²
​

Answer 1

Given that: x+y=12 and xy = 32

$\small \small \small\small\small \small \small\small \bold{Using \: the \: identity, (x+y)^2= x^2 +2xy+y^2}$

$\bold{⇒ (12)^2 =x^2 +64+y^2}$

$\bold{⇒ 144 = x^2 +y^2+64}$

$\bold{⇒ 144−64 = x^2 +y^2}$

$\bold{⇒ 80 = x^2 +y^2}$

$\small \bold{The \: answer \: is \: x^2+y^2 = 80}$

Answer 2

Answer:

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

Step-by-step explanation:

Expansions To be Used :

$(a+b)^{2}=a^{2}+b^2+2ab\\=>(x+y)^{2}=x^{2}+y^2+2xy\\=>12^2=x^{2}+y^2+2(32)\\=>144=x^{2}+y^2+64\\=>144-64=x^{2}+y^2\\=>x^{2}+y^2=80$