Let $a$ and $b$ be the roots of the quadratic equation $2x^2 - 7x + 2 = -x^2 + 4x + 9.$ Find $\frac{1}{a-1}+\frac{1}{b-1}.$
First, we move all the terms to the left-hand side to get
\[3x^2 - 11x -7 = 0.\]We can easily verify that 1 is not a root, so that
\[\frac{1}{a - 1} + \frac{1}{b - 1} = \frac{a + b - 2}{ab - (a + b) + 1}.\]By Vieta's formulas, $a + b = \frac{11}{3}$ and $ab = -\frac{7}{3}.$ Hence,
\[\frac{11/3 - 2}{-7/3 - 11/3 + 1} = \frac{\frac{5}{3}}{\frac{-17}{3}} = \boxed{-\frac{5}{17}}.\]
We can start by simplifying the given equation.
We have $2x^2 - 7x + 2 = -x^2 + 4x + 9.$
Since we have quadratic terms on both sides, we can combine like terms and set the resulting equation to $0.$
This gives us $2x^2 + x + 7 = 0.$
Now, let's find the roots of this quadratic equation using the quadratic formula.
The quadratic formula states that if we have a quadratic equation of the form $ax^2 + bx + c = 0,$ where $a,$ $b,$ and $c$ are constants, then the solutions are given by
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
Applying the quadratic formula to our equation $2x^2 + x + 7 = 0,$ we find
$$x = \frac{-1 \pm \sqrt{1 - 4(2)(7)}}{2(2)}.$$
Simplifying this expression further, we have
$$x = \frac{-1 \pm \sqrt{1 - 56}}{4} = \frac{-1 \pm \sqrt{-55}}{4}.$$
Since the discriminant is negative, the roots are complex conjugates. Therefore, we have
$$x = \frac{-1 \pm i \sqrt{55}}{4}.$$
Let $a$ and $b$ be the roots of the quadratic equation $2x^2 - 7x + 2 = -x^2 + 4x + 9.$ Then $a = \frac{-1 + i \sqrt{55}}{4}$ and $b = \frac{-1 - i \sqrt{55}}{4}.$
We are asked to find $\frac{1}{a-1} + \frac{1}{b-1}.$ Let's compute this expression.
We have
$$\frac{1}{a-1} + \frac{1}{b-1} = \frac{1}{\frac{-1 + i \sqrt{55}}{4} - 1} + \frac{1}{\frac{-1 - i \sqrt{55}}{4} - 1}.$$
To simplify this expression, let's find a common denominator.
The common denominator is $4,$ so we can rewrite the expression as
$$\frac{1}{\frac{-1 + i \sqrt{55}}{4} - 1} + \frac{1}{\frac{-1 - i \sqrt{55}}{4} - 1} = \frac{1}{\frac{-1 + i \sqrt{55}}{4} - \frac{4}{4}} + \frac{1}{\frac{-1 - i \sqrt{55}}{4} - \frac{4}{4}}.$$
Simplifying further, we have
$$\frac{1}{\frac{-1 + i \sqrt{55} - 4}{4}} + \frac{1}{\frac{-1 - i \sqrt{55} - 4}{4}} = \frac{1}{\frac{-5 + i \sqrt{55}}{4}} + \frac{1}{\frac{-5 - i \sqrt{55}}{4}}.$$
Next, we can simplify the expressions inside the denominators.
For the first term, we have
$$\frac{-5 + i \sqrt{55}}{4} = \frac{(-5 + i \sqrt{55})(-5 - i \sqrt{55})}{4(-5 - i \sqrt{55})} = \frac{-25 - 5i \sqrt{55} + 5i \sqrt{55} + 55}{(-20 - 4i \sqrt{55})} = \frac{30}{-20 - 4i \sqrt{55}} = \frac{3}{-2 - i \sqrt{55}}.$$
For the second term, we have
$$\frac{-5 - i \sqrt{55}}{4} = \frac{(-5 - i \sqrt{55})(-5 + i \sqrt{55})}{4(-5 + i \sqrt{55})} = \frac{-25 + 5i \sqrt{55} - 5i \sqrt{55} - 55}{(-20 + 4i \sqrt{55})} = \frac{-80}{-20 + 4i \sqrt{55}} = \frac{4}{1 - i \sqrt{55}}.$$
Putting everything together, we have
$$\frac{1}{a-1} + \frac{1}{b-1} = \frac{3}{-2 - i \sqrt{55}} + \frac{4}{1 - i \sqrt{55}}.$$