a. maths again

I think Guttvennnnnnnnn had made a mistake in Part C

PART A
Since tanα and tanβ are the roots of x^2 + 12x + 6= 0;
Then (x - tanα) (x - tanβ) = 0
==> x^2 - (tanα+tanβ)x +tanα*tanβ
Substitue into the original equation:
tanα + tanβ = -12
tanα*tanβ = 6


Since tanα+tanβ = -12 which is -ve means at least one of the elements of tanα or tanβ is/are negative. ...(a)

tanα*tanβ = 6 which is +ve suggest that either both tanα and tanβ are positive or both tanα and tanβ are negative. ...(b)
To reconcile (a) and (b), I can conclude that both tanα and tanβ are -ve.

Given by the question itself, α and β lie between 0⁰ and 180⁰,
ie, 0⁰≦α, β ≦180⁰ . But if tanα or tanβ lies between 0⁰ and 90⁰, tanα or tanβ would be +ve, from the reconciliation of (a) and (b), α and β must lie between 90⁰ and 180⁰, ie, obtuse angle: 90⁰≦α, β ≦180⁰ (Answer.)


PART B
Since tan(α+β) = (tanα+tanβ)/(1-tanαtanβ)
From PART A,
tan(α+β) = -12 / (1-6) = 12/5 (Answer)


PART C
From PART A, since 90⁰≦α, β ≦180⁰, then 180⁰≦α + β ≦360⁰

From PART B, tan(α+β) = 12/5
By the rule of ASTC *, and I can exclude (α+β) lies between 0⁰ and 90⁰, the only conclusion is (α+β) lies between 180⁰ and 270⁰, ie. in quadrant III

ASTC * , ie. All are +ve if angle lies between 0⁰ and 90⁰, only Sine is +ve if angle lies between 90⁰ and 180⁰, only Tangent is +ve if angle lies between 180⁰ and 270⁰, and only Cosine is +ve if angle lies between 270⁰ and 360⁰,

By drawing a simple triangle, ie a^2 + b^2 = c^2, c^2 = 5^2 +12^2 = 169,
the c = +13 or -13

sin(α+β) = -12/13 Rule of ASTC, must be -ve
cos(α+β) = -5/13 Rule of ASTC, must be -ve

Hence, the requested quartic equation, let y be the unknown:
(y +12/13) (y+5/13) = 0
==> y^2 + 5y/13 + 12y/13 + (12*5) / (13*13) =0
==> y^2 + 17y/13 + 60/169 =0
==> 169y^2 +13*17y +60 = 0
==> 169y^2 + 221y + 60 = 0 (Answer)

From (b), tan(α+β) = 12/5
then, sin(α+β)=12/13
cos(α+β)=5/13
therefore, sum of roots= sin(α+β)+cos(α+β)=17/13
product of roots= sin(α+β)cos(α+β)=60/169
the equation required: x^2 - (17/13)x + (60/169) = 0