Question Number 83127 by jagoll last updated on 28/Feb/20
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{x}\sqrt{\mathrm{3}}\:+\mathrm{y}\: \\ $$$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\:\mathrm{3}\:? \\ $$
Answered by mr W last updated on 28/Feb/20
![let u=(√3)x+y y=u−(√3)x x^2 +(u−(√3)x)^2 −x(u−(√3)x)=3 (4+(√3))x^2 −(1+2(√3))ux+u^2 −3=0 such that x∈R, Δ=(1+2(√3))^2 u^2 −4(4+(√3))(u^2 −3)≥0 −3u^2 +12(4+(√3))≥0 u^2 ≤4(4+(√3)) −2(√(4+(√3)))≤u≤2(√(4+(√3))) i.e. the range of x(√3)+y is [−2(√(4+(√3))),2(√(4+(√3)))]](Q83133.png)
$${let}\:{u}=\sqrt{\mathrm{3}}{x}+{y} \\ $$$${y}={u}−\sqrt{\mathrm{3}}{x} \\ $$$${x}^{\mathrm{2}} +\left({u}−\sqrt{\mathrm{3}}{x}\right)^{\mathrm{2}} −{x}\left({u}−\sqrt{\mathrm{3}}{x}\right)=\mathrm{3} \\ $$$$\left(\mathrm{4}+\sqrt{\mathrm{3}}\right){x}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){ux}+{u}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${such}\:{that}\:{x}\in{R}, \\ $$$$\Delta=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)\left({u}^{\mathrm{2}} −\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$−\mathrm{3}{u}^{\mathrm{2}} +\mathrm{12}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)\geqslant\mathrm{0} \\ $$$${u}^{\mathrm{2}} \leqslant\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right) \\ $$$$−\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}\leqslant{u}\leqslant\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}} \\ $$$${i}.{e}.\:{the}\:{range}\:{of}\:{x}\sqrt{\mathrm{3}}+{y}\:{is}\:\left[−\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}},\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}\right] \\ $$
Commented by jagoll last updated on 28/Feb/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$