Problem
Let [tex]\alpha[/tex] and [tex]\beta[/tex] be the solutions of the quadratic equation [tex]2x^2-6x-7=0[/tex]. Find the value of [tex]\frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha}[/tex].

Respuesta :

Answer:

[tex]-\dfrac{463}{7}[/tex]

Explanation:

Given equation: 2x² - 6x - 7 = 0

In quadratic equation: ax² + bx + c

[tex]Sum \ of \ roots : \alpha + \beta = \dfrac{-b}{a}[/tex]

[tex]product \ of \ roots : \alpha \beta = \dfrac{c}{a}[/tex]

So, here given:

[tex]Sum : \alpha + \beta = \dfrac{-(-6)}{2} = 3[/tex]

[tex]Product : \alpha \beta = \dfrac{-7}{2} = - 3.5[/tex]

For finding value:

[tex]\dfrac{\alpha^3}{\beta } + \dfrac{\beta^3 }{\alpha }[/tex]

join fractions

[tex]\dfrac{\alpha^4+ \beta^4}{\alpha \beta }[/tex]

factor out

[tex]\dfrac{(\alpha^2 + \beta ^2)^2 -2\alpha ^2 \beta ^2 }{\alpha \beta }[/tex]

when factored more

[tex]\dfrac{((\alpha + \beta)^2 -2\alpha \beta )^2 -2(\alpha \beta)^2 }{\alpha \beta }[/tex]

insert values inside

[tex]\dfrac{((3)^2 -2(-3.5) )^2 -2(-3.5)^2 }{-3.5 }[/tex]

calculate for value

[tex]-\dfrac{463}{7}[/tex]