Given equation are
x2+ax+12=0x2+bx+15=0x2+(a+b)x+36=0
Let α be the positive common root of the three equation.
Let other roots be β,γ,δ respectively.
∴α+β=−a, α⋅β=12
α+γ=−b, α⋅γ=15
α+δ=−(a+b), α⋅δ=36
(α+β)+(α+γ)=−(a+b)=α+δ⇒α+β+γ=δ ⋯(1)
Also,
α(β+γ+δ)=12+15+36=63⇒α(2δ−α)=63 [from (1)]⇒2α⋅δ−α2=63⇒72−α2=63⇒α=3 (∵α is positive)
If α=3⇒β=4, γ=5, δ=12
∴α+β+γ+δ=24