Find the values of the parameter `a` such that the rots `alpha,beta` of the equation `2x^2+6x+a=0` satisfy the inequality `alpha//beta+beta//alpha<2.`

The least prime integral value of '2a' such that the roots alpha,beta of the equation 2x^(2)+6x+a=0 satisfy the inequality (alpha)/(beta)+(beta)/(alpha)<2 is

If alpha " and " beta are the roots of the quadratic equation x^2 -5x +6 =0 then find alpha/beta + beta/alpha

If alpha,beta are roots of the equation 2x^(2)+6x+b=0(b<0), then (alpha)/(beta)+(beta)/(alpha) is less than

If alpha and beta are roots of the equation x^2 – x-1 = 0 , then the equation where roots are alpha/beta and beta/alpha is:

If alpha, beta are the roots of equation 3x^(2)-4x+2=0 , then the value of (alpha)/(beta)+(beta)/(alpha) is

If alpha and beta are the roots of equation 2x^(2)-3x+5=0 then find the value of alpha^(2)beta+beta^(2)alpha

If 'alpha,beta' are the zeroes of ^(@)6x^(^^)2+x-2 then evaluate 'alpha/beta + beta/alpha.

If alpha and beta are the roots of the equation 2x^2+3x-6 then find the value of alpha^2+beta^2