The quadratic x^2 + 3/2*x - 1 has the following unexpected property: the roots, which are 1/2 and -2, are one 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 two less than these coefficients.
+9674 Let \(\alpha, \beta\) be the roots.
Then the quadratic is \(x^2 - (\alpha + \beta)x + \alpha \beta\).
Set \(\begin{cases} \alpha = -(\alpha + \beta) - 2\\ \beta = \alpha\beta - 2 \end{cases}\).
Then \(\begin{cases} \alpha = \dfrac{-\beta - 2}2\\ \beta = \alpha \beta - 2 \end{cases}\).
Substituting, we have
\(\beta = \beta\left(\dfrac{-\beta-2}{2}\right) - 2\\ \beta = -\dfrac{\beta^2}2 - \beta - 2\\ \dfrac{\beta^2}2 + 2\beta + 2 = 0\\ \beta^2 + 4\beta + 4 = 0\\ (\beta + 2)^2 = 0\\ \beta = -2\)
Now, \(\alpha = \dfrac{-(-2) - 2}2 = 0\)
The required quadratic is \(x^2 - (0 + (-2))x + 0(-2)\)
Simplifying, the required quadratic is \(x^2 + 2x \)
