MHB What is the value of the sum of reciprocals of the roots in a cubic equation?

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The discussion focuses on finding the value of the sum of the reciprocals of the squares of the roots, specifically expressed as 1/p² + 1/q² + 1/r², for the cubic equations x³ + ax² - 4x + 3 = 0 and x³ + ax - 4x + 3 = 0. Participants are encouraged to solve these problems in terms of the parameter 'a'. The conversation highlights the collaborative effort in solving mathematical challenges, with positive reinforcement for contributions. Overall, the thread emphasizes problem-solving in cubic equations and the relationships between roots and coefficients. Engaging with challenging mathematical problems is encouraged.
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If $p,\,q,\,r$ are roots of the equation $x^3+ax^2-4x+3=0$, find the value of $\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}$ in terms of $a$.
 
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anemone said:
If $p,\,q,\,r$ are roots of the equation $x^3+ax-4x+3=0$, find the value of $\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}$ in terms of $a$.

We have (Vieta's formulas):
\begin{aligned}
p+q+r&=-a \\
pq+pr+qr&=-4 \\
pqr&=-3 \\
\end{aligned}

Define $A=pq, B=pr, C=qr$. Then:
\begin{array}{}
A+B+C&=&-4 \\
AB+AC+BC&=&pqr(p+q+r)&=&-3 \cdot -a &=& 3a \\
A^2+B^2+C^2 &=& (A+B+C)^2 - 2(AB+AC+BC) &=& (-4)^2 - 2\cdot 3a&=&16-6a
\end{array}

Therefore:
$$\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}
=\frac{(pq)^2 + (pr)^2 + (qr)^2}{(pqr)^2}
=\frac{A^2 + B^2 + C^2}{(-3)^2}
=\frac{16-6a}{9}$$
 
anemone said:
If $p,\,q,\,r$ are roots of the equation $x^3+ax-4x+3=0$, find the value of $\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}$ in terms of $a$.

I think you mean $x^3+ax^2-4x+3= 0 $

if p, q ,r are roots of f(x) then 1/p,1/q. 1/r are roots of f(1/x) = 0

or $3x^3-4x^2+ax+1 = 0 $

so $(\frac{1}{p}+\frac{1}{q} + \frac{1}{r}) = 4/3$

$(\frac{1}{pq} + \frac{1}{qr} +\frac{1}{rp} ) = a/3$

or hence $\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} = (\frac{1}{p}+\frac{1}{q} + \frac{1}{r})^2 – 2(\frac{1}{pq} + \frac{1}{qr} +\frac{1}{rp} ) = \frac{16-6a}{9}$
 
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Hey, the two of you are doing so great today by solving my two challenge problems in such a perfect and neat way! :o Well done!(Cool)

I am looking forward to see some challenging problems to play instead...(Wink)(Sun)
 
Hello, anemone!

\text{If }p,\,q,\,r\text{ are roots of the equation}
x^3+ax-4x+3=0,\,\text{find the value}
\text{of }\tfrac{1}{p^2}+\tfrac{1}{q^2}+\tfrac{1}{r^2}\, \text{ in terms of }a.
From Vieta's formulas: . \begin{array}{ccc}p + q + r \:=\:\text{-}a & [1] \\ pq + qr + pr \:=\:\text{-}4 & [2] \\ pqr \:=\:\text{-}3 & [3] \end{array}

Square [2]:
. . (pq+qr+pr)^2 \:=\: (\text{-}4)^2

. . p^2q^2 + q^2r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \:=\:16

. . p^2q^2+ q^2r^2+p^2r^2 + 2\underbrace{pqr}_{\text{-}3}\underbrace{(p+q+r)}_{\text{-}a} \:=\:16

. . p^2q^2+q^2<br /> r^2+p^2r^2 + 6a \:=\:16

. . p^2q^2+q^2r^2+p^2r^2 \:=\:16-6a .[4]Square [3]: .(pqr)^2 \:=\; (\text{-}3)^2 \quad\Rightarrow\quad p^2q^2r^2 \:=\:9 .[5]We have: .\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} \;=\; \frac{p^2q^2 + q^2r^2 + p^2r^2}{p^2q^2r^2}Substitute [4] and [5].

Therefore: .\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} \;=\;\frac{16-6a}{9}
 
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soroban said:
Hello, anemone!

From Vieta's formulas: . \begin{array}{ccc}p + q + r \:=\:\text{-}a &amp; [1] \\ pq + qr + pr \:=\:\text{-}4 &amp; [2] \\ pqr \:=\:\text{-}3 &amp; [3] \end{array}

Square [2]:
. . (pq+qr+pr)^2 \:=\: (\text{-}4)^2

. . p^2q^2 + q^2r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \:=\:16

. . p^2q^2+ q^2r^2+p^2r^2 + 2\underbrace{pqr}_{\text{-}3}\underbrace{(p+q+r)}_{\text{-}a} \:=\:16

. . p^2q^2+q^2<br /> r^2+p^2r^2 + 6a \:=\:16

. . p^2q^2+q^2r^2+p^2r^2 \:=\:16-6a .[4]Square [3]: .(pqr)^2 \:=\; (\text{-}3)^2 \quad\Rightarrow\quad p^2q^2r^2 \:=\:9 .[5]We have: .\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} \;=\; \frac{p^2q^2 + q^2r^2 + p^2r^2}{p^2q^2r^2}Substitute [4] and [5].

Therefore: .\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} \;=\;\frac{16-6a}{9}

Thanks for participating and well done, soroban! :)
 
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