Finding $(\alpha - \gamma)(\alpha - \delta)$ if they are roots of given quadratic equations

Question

If $\alpha, \beta$ are roots of the equation $x^2 + px - q = 0$. $\gamma , \delta$ are roots of equation $x^2 + px -r$, then find the value of $(\alpha - \gamma )(\alpha - \delta)$.

Answer - $q-r$

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My try -

$\alpha + \beta= -p$ and $\alpha \beta = -q$

similarly,

$\gamma + \delta = -p$ and $\gamma \delta = -r$

Then to we've to find:

$(\alpha - \gamma)(\alpha - \delta) = \alpha^2 - \alpha \delta - \alpha \gamma + \gamma \delta $ out of which only $\gamma \delta$ is known, then how to find the rest?

Also, when noticed carefully about the question, we find that question is $(\alpha - \gamma)(\alpha - \delta)$ which doesn't have $\beta$ in its product, which makes the question more confusing.

Thanks in Advance :)

Answer 1

We have $\alpha + \beta = \gamma + \delta \implies \beta = \gamma + \delta - \alpha$

Now, $(\alpha - \gamma)(\alpha - \delta) = \alpha^2 - \alpha \delta - \alpha \gamma + \gamma \delta = \alpha(\alpha - \delta - \gamma) + \gamma\delta = \alpha(-\beta) +\gamma\delta = \gamma\delta - \alpha\beta $

$= -r + q$

Answer 2

We get by solving the quadratic $$x_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+q}$$ and

$$x_{3,4}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}+r}$$ so we get

$$(\alpha-\gamma)(\beta-\delta)=\left(\sqrt{\frac{p^2}{4}+q}-\sqrt{\frac{p^2}{4}+r}\right)\left(\sqrt{\frac{p^2}{4}+q}+\sqrt{\frac{p^2}{4}+r}\right)=\frac{p^2}{4}+q-\frac{p^2}{4}-r=q-r$$ as you stated.

Answer 3

By the given: $$(\alpha-\gamma)(\alpha-\delta)=\alpha^2+p\alpha-r=q-r.$$