Question

If alpha and beta are the zeroes of the polynomial 6x2+x-2 find the value og alpha/beta + beta/alpha

Answer 1

`6x^2+x-2=6x^2+4x-3x-2=2x(3x+2)-1(3x+2)\\\\=(3x+2)(2x-1)\\\\3x+2=0\to x=-(2)/(3)\\\\2x-1=0\to x=(1)/(2)\\\\\alpha=-(2)/(3);\ \beta=(1)/(2)\\\\(\alpha)/(\beta)+(\beta)/(\alpha)=(-(2)/(3))/((1)/(2))+((1)/(2))/(-(2)/(3))=-(2)/(3)\cdot(2)/(1)-(1)/(2)\cdot(3)/(2)=-(4)/(3)-(3)/(4)\\\\=-(16)/(12)-(9)/(12)=-(25)/(12)=-2(1)/(12)`



`use\ Vieta's\ formula:\\\\(\alpha)/(\beta)+(\beta)/(\alpha)=(\alpha^2+\beta^2)/(\alpha\beta)=(\alpha^2+2\alpha\beta+\beta^2-2\alpha\beta)/(\alpha\beta)=((\alpha+\beta)^2-2\alpha\beta)/(\alpha\beta)=((\alpha+\beta)^2)/(\alpha\beta)-2\\\\\alpha+\beta=(-b)/(a);\ \alpha\beta=(c)/(a)\\\\((\alpha+\beta)^2)/(\alpha\beta)-2=(\left((-b)/(a)\right)^2)/((c)/(a))-2=(b^2)/(a^2)\cdot(a)/(c)-2=(b^2)/(ac)-2`


`6x^2+x-2\\\\a=6;\ b=1;\ c=-2\\\\(\alpha)/(\beta)+(\beta)/(\alpha)=(1^2)/(6\cdot(-2))-2=(1)/(-12)-2=-2(1)/(12)`