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The equation $x^2+px+q=0$ has roots $\alpha , \beta$; the equation $y^2+ry+s=0$has roots $\delta, \gamma$. Find $$(\delta-\alpha)(\gamma-\alpha)(\delta-\beta)(\gamma-\beta)$$ as a polynomial of p,q,r,s.( This polynomial is called the resultant of two quadratic polynomials, it is equal to zero if these two polynomials have a common root.)

The question comes from Gelfand and Shen 'Algebra' . It comes after a section on Vieta's Theorem $$\alpha + \beta = -p$$ $$\alpha.\beta = q$$ I have tried multiplying out the brackets but I can't see how to relate the terms to p,q,r and s. I think the problem is designed to be solved with basic algebra.

mikoyan
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  • You could use the ABC formula for $\alpha$, $\beta$, $\gamma$ and $\delta$ and then substitute those values in the required expression. – Ragnar May 01 '14 at 20:33

1 Answers1

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Hint: $$ x^2+px+q=(x-\alpha)(x-\beta) $$ implies $$ (\delta-\alpha)(\delta-\beta)=\delta^2+p\delta+q $$ and $$ (\gamma-\alpha)(\gamma-\beta)=\gamma^2+p\gamma+q. $$ It reduces to compute $$ (\delta^2+p\delta+q)(\gamma^2+p\gamma+q). $$ Expanding it, then applying the other Vieta's Theorem, the result is $$ q^2+s^2-pqr-srp+qr^2+sp^2-2sq. $$

Ma Ming
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