Using Vieta's to create polynomial The quadratic $x^2+\frac{3}{2}x-1$ has the following unexpected property: the roots, which are $\frac{1}{2}$ and $-2$, are one $\textit{less}$ than the final two coefficients. Now find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one $\textit{more}$ than these coefficients.
I thought of using Vieta's and designating roots p and q, but I wasn't able to go far. Here is my thought process. 
Let there be 2 roots, $p$ and $q$. 
$$p+q = -b$$ 
$$pq= a$$ 
$$p=-b+1$$ 
$$q = a+1$$ 
At this point, I got lost. 

Let $x^2 + ax + b$ be the wanted polynomials and $x_1,x_2$ it's roots. Then by Vieta's Formulae we have $x_1 + x_2 = -a$ and $x_1x_2 = b$. Also from the condition $x_1 = a + 1$ and $x_2 = b+1$. 
Now subsituting in the first equation we have: $2a + b = -2$, so we get that $a= \frac{-2-b}{2}$. Then $x_1 = \frac{-2-b}{2} - 1 = -\frac{4+b}{2}$. Substituting in the second Vieta's Formula we have:
$$b = \left(-\frac{4+b}{2}\right)(b-1) \iff -2b = b^2 + 3b - 4 \iff b^2 + 5b - 4 =0 $$
Solving this equation will give you a solution.
