If alpha beta are the roots of x+1/x=10/3,then relation between alpha

1.	If alpha beta are the roots of x+1/x=10/3,then relation between alpha and beeta.
Answer» $\large{\underline{\underline{\red{\sf{GIVEN:}}}}}$ $\tt{\alpha \:and\: \beta \:are\: zeroes \:of \:equ^{n}}$ $\tt{The\:equ^{n}\:is\:x+\dfrac{1}{x}=\dfrac{10}{3}}$ $\large{\underline{\underline{\red{\sf{TO\:FIND:}}}}}$ $\tt{Relation\: between\:\alpha\:and\:\beta}$ $\large{\underline{\underline{\red{\sf{ANSWER:}}}}}$ $\sf{Given \:<klux>QUADRATIC</klux> \:<klux>EQUATION</klux>\: is \:x +\dfrac{ 1}{x }=\dfrac{ 10}{3}}$ . So , firstly let's simplify the equation , $\tt{\implies x+\dfrac{1}{x}=\dfrac{10}{3}}$ $\tt{\implies \dfrac{x^2+1}{x}=\dfrac{10}{3}}$ $\tt{\implies 3(x^2+1)=10x}$ $\tt{\implies 3x^2+3=10x }$ $\tt{\implies 3x^2-10x+3=0}$ Now we CONVERTED this equⁿ in standard FORM of a quadratic equation that is ax²+bx+c . Now we know a relationship between zeroes and COEFFICIENTS as , $\large{\underline{\boxed{\purple{\rm{\leadsto Product\:of\: Zeroes=\dfrac{Coefficient\:of\: constant\:term}{Coefficient\:of\:x^2}=\dfrac{c}{a}}}}}}$ Now here zeroes are alpha and beta. Here , $\orange{\sf{Coefficient\:of\:c\:=\:3}}$ $\orange{\sf{Coefficient\:of\:x^2\:=\:3}}$ $\tt{\implies \alpha\beta=\dfrac{3}{3}}$ $\tt{\implies \alpha\beta=1}$ $\underline{\green{\tt{\underset{\purple{Required\: relationship}}{\underbrace{\dag\alpha=\dfrac{1}{\beta}}}}}}$