Relationship between Roots and Coefficients Polynomial Questions (1 Viewer)

202025 · Jan 25, 2019

Solve the equation 6x^4 - 29x^3 + 40x^2 - 7x - 12 = 0, the product of two of the roots being 2.

Solve the equation 8x^4 - 2x^3 - 27x^2 + 6x + 9 = 0, given that the sum of two of its roots is zero

I really need assistance with these questions. I can form equations with the roots but I am having a lot of trouble solving for the roots.

Help would be greatly appreciated.

yleming · Jan 26, 2019

1) $6x^4 - 29x^3 + 40x^2 - 7x - 12 = 0$

Let the roots be $\alpha, \beta, \gamma$ and $\delta.$

$\alpha\beta = 2 \Rightarrow \alpha = \frac{2}{\beta}$

$\alpha + \beta + \gamma + \delta = -\big( \frac{-29}{6}\big) = \frac{29}{6}$

$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \delta\gamma = \frac{40}{6} = \frac{20}{3}$

$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\big( \frac{-7}{6} \big) = \frac{7}{6}$

$\alpha\beta\gamma\delta = \frac{-12}{6} = -2$

From these equations and the given condition:

$\alpha + \beta + \gamma + \delta = \frac{29}{6} \Rightarrow (1)$

$2+ \big(\alpha+\beta\big)\big(\gamma+\delta\big) + \delta\gamma = \frac{20}{3} \Rightarrow (2)$

$2 \big(\gamma + \delta\big) + \gamma\delta\big( \alpha + \beta \big) = \frac{7}{6} \Rightarrow (3)$

$\gamma\delta = -1 \Rightarrow (4)$

Subbing (4) into (3):

$2 \big( \gamma + \delta \big) - \big( \alpha + \beta \big) = \frac{7}{6}$

$\big( \alpha + \beta \big) = 2 \big( \gamma + \delta \big) - \frac{7}{6} \Rightarrow (5)$

Subbing (5) into (1):

$3 \big( \gamma + \delta \big) = 6$

$\big( \gamma + \delta \big) = 2 \Rightarrow (6)$

Subbing (4) and (6) into (2):

$1+2\big(\alpha+\beta\big) = \frac{20}{3}$

$\big(\alpha+\beta\big) = \frac{17}{6}$

Using the sum and product of $\alpha, \beta$ as well as $\gamma, \delta$ , we can create quadratic equations:

$x^2 - \frac{17}{6}x +2 = 0$

$x^2 - 2x - 1 = 0$

Solving for $x$ in these equations, the roots are 1+√2, 1-√2, 3/2, 4/3.

2) $8x^4 - 2x^3 - 27x^2 + 6x + 9 = 0$

Let the roots be $\alpha, \beta, \gamma$ and $\delta.$

$\alpha + \beta = 0 \Rightarrow \alpha = -\beta$

$\alpha + \beta + \gamma + \delta = -\big( \frac{-2}{8} \big) = \frac{1}{4}$

$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \delta\gamma = -\frac{27}{8}$

$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{6}{8} = -\frac{3}{4}$

$\alpha\beta\gamma\delta = \frac{9}{8}$

From these equations and the given condition:

$\gamma + \delta = \frac{1}{4}\Rightarrow (1)$

$\alpha\beta + \delta\gamma = -\frac{27}{8} \Rightarrow (2)$

$\alpha\beta\gamma + \alpha\beta\delta = \alpha\beta\big(\gamma+\delta\big) = -\frac{3}{4} \Rightarrow (3)$

$\alpha\beta\gamma\delta = \frac{9}{8} \Rightarrow (4)$

Subbing (1) into (3):

$\alpha\beta = -3 \Rightarrow (5)$

Subbing (5) into (4):

$\gamma\delta = -\frac{3}{8}$

Using the sum and product of $\alpha, \beta$ as well as $\gamma, \delta$ , we can create quadratic equations:

$x^2 - 0 - 3 = 0$

$x^2 - \frac{1}{4}x - \frac{3}{8} = 0$

Solving for $x$ in these equations, the roots are √3, -√3, 3/4, -0.5.

Hope this all makes sense!

Relationship between Roots and Coefficients Polynomial Questions (1 Viewer)

202025

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yleming

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