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Roots of a cubic equation

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anemone

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Feb 14, 2012
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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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Klaas van Aarsen

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Mar 5, 2012
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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:
$$\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}
=\frac{(pq)^2 + (pr)^2 + (qr)^2}{(pqr)^2}
=\frac{A^2 + B^2 + C^2}{(-3)^2}
=\frac{16-6a}{9}$$
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
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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anemone

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Feb 14, 2012
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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! :eek: Well done!(Cool)

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

soroban

Well-known member
Feb 2, 2012
409
Hello, anemone!

[tex]\text{If }p,\,q,\,r\text{ are roots of the equation}[/tex]
[tex]x^3+ax-4x+3=0,\,\text{find the value}[/tex]
[tex]\text{of }\tfrac{1}{p^2}+\tfrac{1}{q^2}+\tfrac{1}{r^2}\, \text{ in terms of }a.[/tex]

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

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

. . [tex]p^2q^2 + q^2r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \:=\:16[/tex]

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

. . [tex]p^2q^2+q^2
r^2+p^2r^2 + 6a \:=\:16[/tex]

. . [tex]p^2q^2+q^2r^2+p^2r^2 \:=\:16-6a[/tex] .[4]


Square [3]: .[tex](pqr)^2 \:=\; (\text{-}3)^2 \quad\Rightarrow\quad p^2q^2r^2 \:=\:9[/tex] .[5]


We have: .[tex]\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}[/tex]


Substitute [4] and [5].

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

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Feb 14, 2012
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Hello, anemone!


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

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

. . [tex]p^2q^2 + q^2r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \:=\:16[/tex]

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

. . [tex]p^2q^2+q^2
r^2+p^2r^2 + 6a \:=\:16[/tex]

. . [tex]p^2q^2+q^2r^2+p^2r^2 \:=\:16-6a[/tex] .[4]


Square [3]: .[tex](pqr)^2 \:=\; (\text{-}3)^2 \quad\Rightarrow\quad p^2q^2r^2 \:=\:9[/tex] .[5]


We have: .[tex]\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}[/tex]


Substitute [4] and [5].

Therefore: .[tex]\frac{1}{p^2} + \frac{1}{q^2} + \frac{1}{r^2} \;=\;\frac{16-6a}{9}[/tex]
Thanks for participating and well done, soroban! :)