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 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 less than these coefficients.
Let a, b be the roots.
The the quadratic is x^2 - (a + b) x + ab.
Set a = -(a + b) - 1
b = ab - 1
Substituting, we have
b =b(-b - 2)/2 - 2
b = -b^2/2 - b - 2
b^2/2 + 2b + 2 = 0
b^2 + 4b + 4 = 0
(b + 2)^2 = 0
b = -2
Now, a = 0.
The required quadratic is x^2 - (0 + (-2)) x + 0(-2) = x^2 + 2x.
+36914 (x-0)(x-1)=0 roots = 0 and +1
x^2 -x + 0 coefficients = -1 and 0
