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anemone
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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 $a$.
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 $a$.
[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]
soroban said: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]
The general formula for finding the roots of a cubic equation, also known as the cubic formula, is given by:
x = ∛(q + (q² + r³)½) + ∛(q - (q² + r³)½) - a/3
Where q = (3ac - b²)/9a² and r = (9abc - 27a²d - 2b³)/54a³.
This formula can be used to find the roots of any cubic equation in the form ax³ + bx² + cx + d = 0.
The discriminant of a cubic equation, given by Δ = 18abcd - 4b³d + b²c² - 4ac³ - 27a²d², can tell us whether the equation has one real root or three real roots.
If Δ < 0, the equation has one real root and two complex conjugate roots.
If Δ > 0, the equation has three distinct real roots.
If Δ = 0, the equation has at least two real roots, one of which is a repeated root.
Yes, a cubic equation can have all three roots as real numbers. This happens when the discriminant (Δ) is greater than 0, indicating that the equation has three distinct real roots.
For example, the cubic equation x³ - 6x² + 11x - 6 = 0 has roots of 1, 2, and 3, all of which are real numbers.
The roots of a cubic equation can be graphically represented by plotting the equation on a coordinate plane. The x-intercepts of the graph will represent the roots of the equation, as these are the points where the equation crosses the x-axis.
For example, the graph of the cubic equation x³ - 6x² + 11x - 6 = 0 has x-intercepts at x = 1, x = 2, and x = 3, which are the roots of the equation.
The relationship between the roots of a cubic equation and its coefficients can be seen in Vieta's formulas. These formulas state that the sum of the roots is equal to the opposite of the coefficient of the x² term, the sum of the products of the roots taken two at a time is equal to the coefficient of the x term, and the product of the roots is equal to the constant term.
For example, in the equation x³ - 6x² + 11x - 6 = 0, the sum of the roots is -(-6) = 6, the sum of the products of the roots taken two at a time is -11, and the product of the roots is -(-6) = 6.