CAIE FP1 2011 June — Question 2

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
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TopicRoots of polynomials

2 The roots of the equation $$x ^ { 3 } + p x ^ { 2 } + q x + r = 0$$ are \(\frac { \beta } { k } , \beta , k \beta\), where \(p , q , r , k\) and \(\beta\) are non-zero real constants. Show that \(\beta = - \frac { q } { p }\). Deduce that \(r p ^ { 3 } = q ^ { 3 }\).

2 The roots of the equation

$$x ^ { 3 } + p x ^ { 2 } + q x + r = 0$$

are $\frac { \beta } { k } , \beta , k \beta$, where $p , q , r , k$ and $\beta$ are non-zero real constants. Show that $\beta = - \frac { q } { p }$.

Deduce that $r p ^ { 3 } = q ^ { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q2}}