AP EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma\) are the roots of the equation \(\mathrm{x}^3+\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\), then \((\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=\)
- A \(\mathrm{p}-\mathrm{qr}\)
- B \(q-r p\)
- C \(\mathrm{r}-\mathrm{pq}\)
- D \(\mathrm{r}+\mathrm{pq}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{r}-\mathrm{pq}\)
Step-by-step Solution
Detailed explanation
\(\alpha+\beta+\gamma = -p\) \((\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (-p-\gamma)(-p-\alpha)(-p-\beta)\) \(= -(p+\alpha)(p+\beta)(p+\gamma)\) Let \(g(y) = (y+\alpha)(y+\beta)(y+\gamma)\). Then \(g(y) = y^3 - py^2 + qy - r\).…
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