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.
+9675 Let α,β be the roots.
Then the quadratic is x2−(α+β)x+αβ.
Set {α=−(α+β)−2β=αβ−2.
Then {α=−β−22β=αβ−2.
Substituting, we have
β=β(−β−22)−2β=−β22−β−2β22+2β+2=0β2+4β+4=0(β+2)2=0β=−2
Now, α=−(−2)−22=0
The required quadratic is x2−(0+(−2))x+0(−2)
Simplifying, the required quadratic is x2+2x
