∴(2x+y)^2-3xy= 1
Let t=2x+y and y=t-2x.
∴t^2-3(t-2x)x= 1
That is 6x 2-3tx+t 2- 1 = 0.
∴△=9t^2-24(t^2- 1)=- 15t^2+24≥0
Solution -2 radical 10/5≤t≤2 radical 10/5.
What is the maximum value of 2x+y? 2 roots 10/5