Answer on Question #45754 – Math - Algebra
a) If the sum of the roots of a cubic equation is 3, the sum of the squares of the roots is 11 and the sum of the cubes of the roots is 27, find the equation. Also, solve the equation and find all its roots.
b) Solve the equation x 3 + 2 x 2 − 2 x − 1 = 0 x^{3} + 2x^{2} - 2x - 1 = 0 x 3 + 2 x 2 − 2 x − 1 = 0
c) Write the systems obtained in 6(a) and 6(b) in matrix form.
Solution
a) We have a cubic equation
x 3 + a x 2 + b x + c = 0. x^{3} + a x^{2} + b x + c = 0. x 3 + a x 2 + b x + c = 0.
Let the roots be denoted α , β , γ \alpha, \beta, \gamma α , β , γ
We know that α + β + γ = 3 \alpha + \beta + \gamma = 3 α + β + γ = 3 α 2 + β 2 + γ 2 = 11 \alpha^2 + \beta^2 + \gamma^2 = 11 α 2 + β 2 + γ 2 = 11 α 3 + β 3 + γ 3 = 27 \alpha^3 + \beta^3 + \gamma^3 = 27 α 3 + β 3 + γ 3 = 27
then by Vieta's formula α + β + γ = − a = 3 \alpha + \beta + \gamma = -a = 3 α + β + γ = − a = 3 ( α β + β γ + γ α ) = b (\alpha\beta + \beta\gamma + \gamma\alpha) = b ( α β + β γ + γ α ) = b α β γ = c \alpha\beta\gamma = c α β γ = c
α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) → → 2 b = 2 ( α β + β γ + γ α ) = ( α + β + γ ) 2 − ( α 2 + β 2 + γ 2 ) = 3 2 − 11 = − 2. \begin{array}{l}
\alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \rightarrow \\
\rightarrow 2b = 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (\alpha + \beta + \gamma)^{2} - (\alpha^{2} + \beta^{2} + \gamma^{2}) = 3^{2} - 11 = -2.
\end{array} α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) → → 2 b = 2 ( α β + β γ + γ α ) = ( α + β + γ ) 2 − ( α 2 + β 2 + γ 2 ) = 3 2 − 11 = − 2.
Then
a = − 3 , b = − 1. a = -3, b = -1. a = − 3 , b = − 1.
Substituting α , β , γ \alpha, \beta, \gamma α , β , γ
( α 3 + β 3 + γ 3 ) − 3 ( α 2 + β 2 + γ 2 ) − ( α + β + γ ) + 3 c = 0 → 27 − 3 ⋅ 11 − 3 + 3 c = 0 → c = 3. (\alpha^{3} + \beta^{3} + \gamma^{3}) - 3(\alpha^{2} + \beta^{2} + \gamma^{2}) - (\alpha + \beta + \gamma) + 3c = 0 \rightarrow 27 - 3 \cdot 11 - 3 + 3c = 0 \rightarrow c = 3. ( α 3 + β 3 + γ 3 ) − 3 ( α 2 + β 2 + γ 2 ) − ( α + β + γ ) + 3 c = 0 → 27 − 3 ⋅ 11 − 3 + 3 c = 0 → c = 3.
So the original equation is
x 3 − 3 x 2 − x + 3 = 0. x 2 ( x − 3 ) − ( x − 3 ) = ( x − 3 ) ( x 2 − 1 ) = ( x − 3 ) ( x − 1 ) ( x + 1 ) = 0. \begin{array}{l}
x^{3} - 3x^{2} - x + 3 = 0. \\
x^{2}(x - 3) - (x - 3) = (x - 3)(x^{2} - 1) = (x - 3)(x - 1)(x + 1) = 0.
\end{array} x 3 − 3 x 2 − x + 3 = 0. x 2 ( x − 3 ) − ( x − 3 ) = ( x − 3 ) ( x 2 − 1 ) = ( x − 3 ) ( x − 1 ) ( x + 1 ) = 0.
Since the roots of the equation x 3 − 3 x 2 − x + 3 = 0 x^{3} - 3x^{2} - x + 3 = 0 x 3 − 3 x 2 − x + 3 = 0
Answer: x 3 − 3 x 2 − x + 3 = 0 x^{3} - 3x^{2} - x + 3 = 0 x 3 − 3 x 2 − x + 3 = 0
b) x 3 + 2 x 2 − 2 x − 1 = 0 x^{3} + 2x^{2} - 2x - 1 = 0 x 3 + 2 x 2 − 2 x − 1 = 0
x = t − 2 3 → t 3 + p x + q = 0 , x = t - \frac{2}{3} \rightarrow t^{3} + p x + q = 0, x = t − 3 2 → t 3 + p x + q = 0 ,
where p = − 2 − 2 2 3 = − 10 3 p = -2 - \frac{2^2}{3} = -\frac{10}{3} p = − 2 − 3 2 2 = − 3 10 q = − 1 + 2 ⋅ 2 3 − 9 ⋅ 2 ( − 2 ) 27 = 25 27 q = -1 + \frac{2 \cdot 2^3 - 9 \cdot 2(-2)}{27} = \frac{25}{27} q = − 1 + 27 2 ⋅ 2 3 − 9 ⋅ 2 ( − 2 ) = 27 25
x 1 = − 3 25 27 ⋅ 2 + ( 25 27 ⋅ 2 ) 2 + ( − 10 3 ) 3 27 + − 3 25 27 ⋅ 2 − ( 25 27 ⋅ 2 ) 2 + ( − 10 3 ) 3 27 . x_{1} = \sqrt[3] -\frac{25}{27 \cdot 2} + \sqrt{(\frac{25}{27 \cdot 2})^{2} + \frac{(-\frac{10}{3})^{3}}{27}} + \sqrt[3] -\frac{25}{27 \cdot 2} - \sqrt{(\frac{25}{27 \cdot 2})^{2} + \frac{(-\frac{10}{3})^{3}}{27}}. x 1 = 3 − 27 ⋅ 2 25 + ( 27 ⋅ 2 25 ) 2 + 27 ( − 3 10 ) 3 + 3 − 27 ⋅ 2 25 − ( 27 ⋅ 2 25 ) 2 + 27 ( − 3 10 ) 3 .
In this case ( 25 27 ⋅ 2 ) 2 + ( − 10 3 ) 3 27 < 0 \left(\frac{25}{27 \cdot 2}\right)^{2} + \frac{\left(-\frac{10}{3}\right)^{3}}{27} < 0 ( 27 ⋅ 2 25 ) 2 + 27 ( − 3 10 ) 3 < 0
That's why the Cardano's method cannot be applied to this problem.
But it is easy to see that x = 1 x = 1 x = 1
x 3 + 2 x 2 − 2 x − 1 = ( x − 1 ) ( x 2 + 3 x + 1 ) = 0. x ^ {3} + 2 x ^ {2} - 2 x - 1 = (x - 1) \left(x ^ {2} + 3 x + 1\right) = 0. x 3 + 2 x 2 − 2 x − 1 = ( x − 1 ) ( x 2 + 3 x + 1 ) = 0.
We have
( x 2 + 3 x + 1 ) = 0 → x 1 , 2 = − 3 ± 3 2 − 4 2 . (x ^ {2} + 3 x + 1) = 0 \rightarrow x _ {1, 2} = \frac {- 3 \pm \sqrt {3 ^ {2} - 4}}{2}. ( x 2 + 3 x + 1 ) = 0 → x 1 , 2 = 2 − 3 ± 3 2 − 4 .
The roots of equation are 1; − 3 + 5 2 ; − 3 − 5 2 \frac{-3 + \sqrt{5}}{2}; \frac{-3 - \sqrt{5}}{2} 2 − 3 + 5 ; 2 − 3 − 5
c) 6(a):
{ x − 3 y + 4 z = 9 4 x + 3 y + 2 z = 7 y − 2 x = 5 − 10 z → { x − 3 y + 4 z = 9 4 x + 3 y + 2 z = 7 − 2 x + y + 10 z = 5 . \left\{ \begin{array}{l} x - 3 y + 4 z = 9 \\ 4 x + 3 y + 2 z = 7 \\ y - 2 x = 5 - 10 z \end{array} \right. \to \left\{ \begin{array}{l} x - 3 y + 4 z = 9 \\ 4 x + 3 y + 2 z = 7 \\ - 2 x + y + 10 z = 5 \end{array} \right.. ⎩ ⎨ ⎧ x − 3 y + 4 z = 9 4 x + 3 y + 2 z = 7 y − 2 x = 5 − 10 z → ⎩ ⎨ ⎧ x − 3 y + 4 z = 9 4 x + 3 y + 2 z = 7 − 2 x + y + 10 z = 5 .
The matrix form is
( 1 − 3 4 4 3 2 − 2 1 10 ) ( x y z ) = ( 9 7 5 ) . \left( \begin{array}{c c c} 1 & - 3 & 4 \\ 4 & 3 & 2 \\ - 2 & 1 & 10 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 9 \\ 7 \\ 5 \end{array} \right). ⎝ ⎛ 1 4 − 2 − 3 3 1 4 2 10 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 9 7 5 ⎠ ⎞ .
6(b):
{ 12 x + 8 y = 440 5 x + 3 y = 175 \left\{ \begin{array}{l} 12 x + 8 y = 440 \\ 5 x + 3 y = 175 \end{array} \right. { 12 x + 8 y = 440 5 x + 3 y = 175
The matrix form is
( 12 8 5 3 ) ( x y ) = ( 440 175 ) . \left( \begin{array}{c c} 12 & 8 \\ 5 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 440 \\ 175 \end{array} \right). ( 12 5 8 3 ) ( x y ) = ( 440 175 ) .
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#340153 on Dec 2023