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Question Number 161439 by amin96 last updated on 17/Dec/21

Commented by bobhans last updated on 18/Dec/21

(Q2) y= ((x−ψ)/(x^2 −5x+6)) ⇒yx^2 −5yx+6y=x−ψ yx^2 −(5y+1)x+(6y+ψ)=0 △≥0 ⇒25y^2 +10y+1−4y(6y+ψ)≥0 y^2 +(10−4ψ)y+1≥0 △<0⇒ 100−80ψ+16ψ^2 −4<0 16ψ^2 −80ψ+96<0 ψ^2 −5ψ+6<0 (ψ−3)(ψ−2)<0 ⇒ 2<ψ<3

$$\left(\mathrm{Q2}\right)\:\mathrm{y}=\:\frac{\mathrm{x}−\psi}{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}\:\Rightarrow\mathrm{yx}^{\mathrm{2}} −\mathrm{5yx}+\mathrm{6y}=\mathrm{x}−\psi \\ $$$$\:\:\mathrm{yx}^{\mathrm{2}} −\left(\mathrm{5y}+\mathrm{1}\right)\mathrm{x}+\left(\mathrm{6y}+\psi\right)=\mathrm{0} \\ $$$$\:\bigtriangleup\geqslant\mathrm{0}\:\Rightarrow\mathrm{25y}^{\mathrm{2}} +\mathrm{10y}+\mathrm{1}−\mathrm{4y}\left(\mathrm{6y}+\psi\right)\geqslant\mathrm{0} \\ $$$$\:\:\mathrm{y}^{\mathrm{2}} +\left(\mathrm{10}−\mathrm{4}\psi\right)\mathrm{y}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\:\bigtriangleup<\mathrm{0}\Rightarrow\:\mathrm{100}−\mathrm{80}\psi+\mathrm{16}\psi^{\mathrm{2}} −\mathrm{4}<\mathrm{0} \\ $$$$\:\mathrm{16}\psi^{\mathrm{2}} −\mathrm{80}\psi+\mathrm{96}<\mathrm{0} \\ $$$$\:\psi^{\mathrm{2}} −\mathrm{5}\psi+\mathrm{6}<\mathrm{0} \\ $$$$\:\left(\psi−\mathrm{3}\right)\left(\psi−\mathrm{2}\right)<\mathrm{0}\:\Rightarrow\:\mathrm{2}<\psi<\mathrm{3}\: \\ $$

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