$f(x)$ is a rational function of the form $f(x)=\frac{ax^2+bx+c}{dx+e}$ and the function $g(x)$ is given by $g(x)=\frac{4}{f(x)}$. It is known that $f(x)$ has an oblique asymptote $y=x+1$ and $g(x)$ has vertical asymptotes $x=\frac{1 \pm \sqrt3}{2} $. Find all the values of $x$ for which $f(x)=g(x)$.
The answer is $x=\frac{-1\pm\sqrt19}{2}$.
Vertical asymptotes of $g(x)$ are when $f(x)=0$ which is when $ax^2+bx+c=0$
Therefore
$ax^2+bx+c=(x-(\frac{1+\sqrt3}{2}))(x-(\frac{1- \sqrt 3}{2}))=0$
$2x^2-2x-1=0$
$f(x)=\frac{2x^2-2x-1}{dx+e}=\frac{2}{d}x-\frac{2}{d}(1+\frac{e}{d})+\frac{1+\frac{2e}{d}+\frac{2e^2}{d^2}}{dx+e}$
$\lim_{x\to0}f(x)=\frac{2}{d}x-\frac{2}{d}(1+\frac{e}{d})=2x+1$
Therefore $d=2$ and $e=-4$
$f(x)=\frac{2x^2-2x-1}{2x-4}$
$g(x)=\frac{4(2x-4)}{2x^2-2x-1}$
$f(x)=g(x)$
$\sqrt((2x^-2x-1)^2)=\sqrt(4(2x-4)^2)$
$2x^2-2x-1=2(2x-4)$
$2x^2-6x+7$
which gives no solutions
I didn't think that method would work as it does not take into account the $\pm$ of the square root.
Expanding gave the equation
$4x^4-8x^3-16x^2+68x-63=0$
which gives 4 solutions,
$x = \frac{-1\pm\sqrt19}{2}$, and the two complex solutions $\frac{3\pm\sqrt5 i}{2}$
which actually gives the correct answer
So, you've got $a,b,c,e$ and $d$, so I'll take it from here.
Instead of writing $$f(x) = g(x)$$
I will write $$f(x) = \frac{4}{f(x)}$$
$$f(x)^2 = 4 $$
Then $$f(x) = ± 2$$
Taking $f(x) = 2$ gives: $2x^2 - 6x + 7 = 0$, which has no real solutions.
Taking $f(x) = -2$ gives: $2x^2 + 2x - 9 = 0$, which has the two aforementioned solutions.
Hope this helps!