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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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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 $a$.
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 $a$.
[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]
Thanks for participating and well done, soroban!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]