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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.

Feb 7, 2021

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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 $$x^2 + 2x$$

Feb 8, 2021