Question

If x² + y² + xy = 1 and x + y = 2, then xy =

(a) –3

(b) 3

(c) $-\frac{3}{2}$

(d) 0

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ {x}^2+y^2+xy=1...\left(1\right) \\ x+y=2...\left(2\right) \\ \mathrm{Now}, \\ x+y=2 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(x+y\right)}^2=2^2 \\ \Rightarrow x^2+y^2+2xy=4\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a+b\right)}^2=a^2+b^2+2ab\right) \\ \Rightarrow x^2+y^2+xy+xy=4 \\ \Rightarrow 1+xy=4\left(\mathrm{From}\left(1\right)\right) \\ \Rightarrow xy=4-1 \\ \Rightarrow xy=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{b}\right). \end{gathered}$$