Find Range Algebraically

[nycmath]

I need a step by step explanation.

Find the range algebraically.

h(x) = (sqrt{x^2 - 4)}/(x - 4)

[Eigenvalue]

1. Let h(x)=y and solve for y to find the range
y(x-4)=x²-4
yx-4y=x²-4
x²-4-yx+4y=0

2. Use the quadratic formula
x=[tex]\frac{y \pm \sqrt{y²-4(4y-4)}}{2}[/tex]
x=[tex]\frac{y \pm \sqrt{x²-16y+16} }{2}[/tex]

3. To have real solutions,the discriminant must be greater than or equal to 0
y²-16y+16[tex]\ge[/tex]0
y=[tex]\frac{16 \pm \sqrt{256-(4*16)} }{2}[/tex]
y=[tex]\frac{16 \pm8 \sqrt{3} }{2}[/tex]=8[tex]\pm[/tex]4[tex]\sqrt{3}[/tex]

As it is greater than or equal to 0, the range is y>8+4[tex]\sqrt{3}[/tex] and y<8-4[tex]\sqrt{3}[/tex]

[nycmath]

1. Let h(x)=y and solve for y to find the range
y(x-4)=x²-4
yx-4y=x²-4
x²-4-yx+4y=0

2. Use the quadratic formula
x=[tex]\frac{y \pm \sqrt{y²-4(4y-4)}}{2}[/tex]
x=[tex]\frac{y \pm \sqrt{x²-16y+16} }{2}[/tex]

3. To have real solutions,the discriminant must be greater than or equal to 0
y²-16y+16[tex]\ge[/tex]0
y=[tex]\frac{16 \pm \sqrt{256-(4*16)} }{2}[/tex]
y=[tex]\frac{16 \pm8 \sqrt{3} }{2}[/tex]=8[tex]\pm[/tex]4[tex]\sqrt{3}[/tex]

As it is greater than or equal to 0, the range is y>8+4[tex]\sqrt{3}[/tex] and y<8-4[tex]\sqrt{3}[/tex]

I'm impressed with your ability to always find a way.