Roots of the Equation!

Algebra Level 3

Let $\alpha$ and $\beta$ be the zeros of $x^{2} + px + 1$ .

Let $\gamma$ and $\delta$ be the zeros of $x^{2} + qx + 1$ .

Find the value of $(\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta)$ in terms of $p$ and $q$ .

q^{2}

p^{2}

p^{2} - q^{2}

q^{2} - p^{2}

1 solution

Tom Engelsman
Mar 4, 2017

Let us write:

$(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta = x^2 + px + 1 = 0$

$(x-\gamma)(x-\delta) = x^2 - (\gamma + \delta)x + \gamma\delta = x^2 + qx + 1 = 0$

which yields $\alpha + \beta = -p; \gamma + \delta = -q; \alpha\beta = \gamma\delta = 1.$ Now, let's expand:

$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta);$

or $[\alpha\beta - \gamma(\alpha+\beta) + \gamma^{2}][\alpha\beta + \delta(\alpha+\beta) + \delta^{2}];$ ;

or $(1 + p\gamma + \gamma^{2})(1 - p\delta + \delta^{2});$

or $(p\gamma - q\gamma)(-p\delta - q\delta)$ ;

or $-\gamma(q - p) \cdot -\delta(q + p)$ ;

or $\gamma\delta(q^2 - p^2)$ ;

or $(1)(q^2 - p^2)$ ;

or $\boxed{q^2 - p^2}.$