Question

dacă x+y=4 şi x^2+y^2=12
calculaţi x-y

Answer 1

2xy = (x+y)^2 - x^2 - y^2 = 16-12 = 4

xy = 2

x=4-y
y*(4-y) = 2
4y-y^2-2 = 0
y^2-4y+2 = 0
Δ = 16 - 8 = 8 Deci $\sqrt{Δ}$ = 2$\sqrt{2}$

y = $\frac{4+2 \sqrt{2} }{2}$ = 2+$\sqrt{2}$
x=4 - 2-$\sqrt{2}$ = 2-$\sqrt{2}$

y=2-$\sqrt{2}$
x=4 - (2-$\sqrt{2}$ )= 2+$\sqrt{2}$

Answer 2

(x+y)²=x²+y²+2xy ⇒4²=12+2xy⇒16-12=2xy⇒4=2xy⇒xy=2
(x-y)²=x²+y²-2xy ⇒(x-y)²=12-2*2⇒(x-y)²=8 ⇒x-y=√8=2√2 sau x-y=-√8=-2√2