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. 
 A: As you said, the key is to use Vieta's formulae.
So, if $p,q$ are the roots of your quadratic polinomial then the first coefficient is $-(p+q)$ and the second is $pq$. You then have to solve
$\begin{cases}
-(p+q)+1 = q \\
pq + 1 = p
\end{cases}$
i guess that the solution of this system is not so hard as you thought
A: Assuming your quadratic is 
$$ x^2 + bx + a $$
with roots $p,q$, Vieta's relationship is as follows
$$ \begin{aligned} p + q &= -b \\ pq &= a \end{aligned} $$
You also have that
$$ \begin{aligned} p &= b + 1 \\ q &= a + 1 \end{aligned} $$
Plugging these in two the first two equations we get
$$ \begin{aligned} (a+1) + (b+1) &= -b \\ (a+1)(b+1) &= a \end{aligned} $$
Simplifying
$$ \begin{aligned} a + 2b + 2 &= 0 \\ ab + b + 1 &= 0 \end{aligned} $$
You can solve this system of equations for $a$ and $b$
A: 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.
A: Note that if you want $p=a+r, q=b+r$ this gives $-a-b=a+r$ or $2a=-(b+r)=-ab$
So for any constant difference whatever, you have $b=-2$ as one of the roots.
