Find Range Algebraically

[nycmath]

See attachment.

[Eigenvalue]

To find the range, solve for x as a function of y:
y(x-4)(x+4)=x(x-2)
y(x²-16)=x²-2x
(y-1)x²+2x-16y=0

Use the quadratic formula:
x=[tex]\frac{-2± \sqrt{4-64y(y-1)} }{2(y-1)}[/tex]
Set the discriminant to be greater than 0, so x will have 2 real solutions

4-64y(y-1)[tex]\ge[/tex]0
-64y²+64y+4[tex]\ge[/tex]0
-16y²+16y+1[tex]\ge[/tex]0
16y²-16y-1[tex]\le[/tex]0

Use the quadratic formula to find y:
y=[tex]\frac{16± \sqrt{16² +4(16)(1)} }{32}[/tex]
=[tex]\frac{16± \sqrt{320} }{32}[/tex]
=[tex]\frac{2± \sqrt{5} }{4}[/tex]

The range is [[tex]\frac{ 2-\sqrt{5} }{4}[/tex], [tex]\frac{2+ \sqrt{5} }{4}[/tex]]

[nycmath]

To find the range, solve for x as a function of y:
y(x-4)(x+4)=x(x-2)
y(x²-16)=x²-2x
(y-1)x²+2x-16y=0

Use the quadratic formula:
x=[tex]\frac{-2± \sqrt{4-64y(y-1)} }{2(y-1)}[/tex]
Set the discriminant to be greater than 0, so x will have 2 real solutions

4-64y(y-1)[tex]\ge[/tex]0
-64y²+64y+4[tex]\ge[/tex]0
-16y²+16y+1[tex]\ge[/tex]0
16y²-16y-1[tex]\le[/tex]0

Use the quadratic formula to find y:
y=[tex]\frac{16± \sqrt{16² +4(16)(1)} }{32}[/tex]
=[tex]\frac{16± \sqrt{320} }{32}[/tex]
=[tex]\frac{2± \sqrt{5} }{4}[/tex]

The range is [[tex]\frac{ 2-\sqrt{5} }{4}[/tex], [tex]\frac{2+ \sqrt{5} }{4}[/tex]]

I like find the range algebraically more than using the graph.