How do you find the zeros of $f (x) = \frac{x^{2} + 6 x + 9}{x^{2} - 9}$ $f(x)=(x^2+6x+9)/(x^2-9)$ ?

Question

4772 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T16:55:23

$x = - 3$ $x=-3$

To find the zeros (roots or solutions) set $y = 0$ $y=0$ , in this case $f (x) = 0$ $f(x)=0$ and solve for $x$ $x$ :

$f (x) = \frac{x^{2} + 6 x + 9}{x^{2} - 9}$ $f(x)=(x^2+6x+9)/(x^2-9)$

$0 = \frac{x^{2} + 6 x + 9}{x^{2} - 9}$ $0=(x^2+6x+9)/(x^2-9)$

$(x^{2} - 9) \cdot 0 = (x^{2} + 6 x + 9)$ $(x^2-9)*0=(x^2+6x+9)$

$0 = x^{2} + 6 x + 9$ $0=x^2+6x+9$

factor:

$(x + 3) (x + 3) = 0$ $(x+3)(x+3) =0$

${(x + 3)}^{2} = 0$ $(x+3)^2 =0$

The function has a double solution at $x = - 3$ $x=-3$

graph{(x^2+6x+9)/(x^2-9) [-15.33, 24.67, -7.52, 12.48]}

How do you find the zeros of f(x)=x2+6x+9x2−9f(x)=(x^2+6x+9)/(x^2-9)?