solve for a 3x^2 +ax -(a^2 -1) = 0 so that all solutions to the equation are real and positive

comparing with the euation

Ax^2 + Bx + C ( as a is there in given equation)

we have

roots = (- B +/- sqrt(B^2-4AC))/ (2A)

and we take the - sign for the second one as if - sign gives positive then + sign being greater adding -B gives positive.

so -B - sqrt(B^2 - 4AC) > 0

or B < - sqrt(B^2-4AC)

put B = a , A = 3 and C = a^2-1


to get a < - sqrt(a^2 + 12(a^1-1)
< - sqrt(13a^2- 12)

but a is -ve and so negate both sides to get

- a > sqrt(13a^2-12)

or a^2 > 13a^2-12 or a^2 < 1 or -1 < a < 1 but as a is -ve -1 < a ..1

as B^2 >= 4AC so a^2 > - 12(a^2- 12) or 13a^2 > 12 or a^2 > (12/13) so a < - sqrt(12/13) ..2

so a = - sqrt(12/13) is the value