### Instruction: The quadratic $x^2 + \frac{3}{2} x - 1$ has the following property: the roots, which are $\frac{1}{2}$ and $-2,$ are one less than the final two coefficients.  Find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one more than these coefficients.  Enter the roots of this quadratic.### Response: To find such a quadratic, I need to think of two numbers that are one more than some non-zero coefficients, and also satisfy the quadratic formula.
Let me call these numbers $r$ and $s,$ and the coefficients $a$ and $b.$
Then I have $r = a + 1$ and $s = b + 1,$ and also $r$ and $s$ are the solutions to $x^2 + ax + b = 0.$
Using the quadratic formula, I get that $r = \frac{-a + \sqrt{a^2 - 4b}}{2}$ and $s = \frac{-a - \sqrt{a^2 - 4b}}{2}.$
Simplifying these equations, I get $a + 1 = \frac{-a + \sqrt{a^2 - 4b}}{2}$ and $b + 1 = \frac{-a - \sqrt{a^2 - 4b}}{2}.$
Adding these equations, I get $a + b + 2 = -a,$ so $2a + b = -2.$
Subtracting these equations, I get $a - b = \sqrt{a^2 - 4b}.$
Squaring both sides, I get $a^2 - 2ab + b^2 = a^2 - 4b.$
Simplifying, I get $-2ab + 4b + b^2 = 0.$
Factoring, I get $b(-2a + b + 4) = 0.$
Since I want $b$ to be non-zero, I can ignore the possibility that $b = 0.$
So I have $-2a + b + 4 = 0,$ or $b = 2a - 4.$
Substituting this into $2a + b = -2,$ I get $2a + 2a - 4 = -2,$ or $4a = 2.$
So $a = \frac{1}{2}.$
Then $b = 2a - 4 = 2 \cdot \frac{1}{2} - 4 = -3.$
So I have found the coefficients $a$ and $b$ that satisfy the given conditions.
The quadratic is $x^2 + \frac{1}{2} x - 3.$
The roots are $r = a + 1 = \frac{3}{2}$ and $s = b + 1 = -2.$