Suppose $a$ and $b$ are real numbers with $ab \neq 0$ and the equation $120a^{2}-120a+1=0$ and $b^{2}-120b+120=0$ hold. Find the value of $\dfrac{1+b+ab}{a}$.
I can solve this problem by putting values of $a$ and $b$ from $120a^{2}-120a+1=0$ and $b^{2}-120b+120=0$ in $\dfrac{1+b+ab}{a}$, but I need some shortcut or easy methos of solve instead of just putting the values obtain.
On
From the first equation:
$120a^2 - 120a + 1 =0 \Rightarrow a = \frac{1}{60} (30 \pm \sqrt{870})$
From the second equation:
$b^2 - 120b + 120 = 0 \Rightarrow b = 2 \left(30 \pm \sqrt{870}\right)$,
just by using the quadratic formula.
Then, you have four possible answers for: $(1 + b + ab) / ab$ given these possible pairings of $a$ and $b$ above.
Hint :
$$120a^2-120a+1=0\iff \left(\frac 1a\right)^2-120\cdot \frac 1a+120=0$$ Hence, $b,\frac 1a$ are the solutions of $$t^2-120t+120=0$$