Find Range Algebraically

[nycmath]

See attachment.

[Guest]

Hi, it's Bob... This question is very interesting.

[tex]y(x-4)=x^2-4x+3 \Rightarrow x^2-(4+y)x+3+4y=0[/tex]

For real [tex]x[/tex] the discriminant must be [tex]\ge 0[/tex]:

[tex](4+y)^2-4(3+4y)\ge 0 \Rightarrow y^2-8y+4\ge 0 \Rightarrow y\le 4-2\sqrt{3}[/tex] or [tex]y\ge 4+2\sqrt{3}[/tex]

([tex]x=4[/tex] is never a root, because it gives [tex]3=0[/tex].)

Range: [tex](-\infty,\,4-2\sqrt{3}]\cup[4+2\sqrt{3},\,+\infty)[/tex]

[nycmath]

Hi, it's Bob... This question is very interesting.

[tex]y(x-4)=x^2-4x+3 \Rightarrow x^2-(4+y)x+3+4y=0[/tex]

For real [tex]x[/tex] the discriminant must be [tex]\ge 0[/tex]:

[tex](4+y)^2-4(3+4y)\ge 0 \Rightarrow y^2-8y+4\ge 0 \Rightarrow y\le 4-2\sqrt{3}[/tex] or [tex]y\ge 4+2\sqrt{3}[/tex]

([tex]x=4[/tex] is never a root, because it gives [tex]3=0[/tex].)

Range: [tex](-\infty,\,4-2\sqrt{3}]\cup[4+2\sqrt{3},\,+\infty)[/tex]

Bingo!