F.4 MATHS quadratic equation)

Let k be a constant if α and β are the roots of the equations 3x^2+2x-k = 0 , find the value of 3α^2β in terms of k

If ax^2 +bx + c = 0, α + β = -b/a, α β = c/a
α + β = -2/3
α β = -k/3
β = -k/(3α)α -k/(3α) = -2/33α α + -k = -2α 3α^2 +2α -k = 0 α = {-2+[2^2-4(3)(-k)]^0.5}/2(3)α = {-2+[4 +12k]^0.5}/6α = {-2+2[1 +3k]^0.5}/6α = {-1+[1 +3k)]^0.5}/3 3α^2β = 3 α (α β) =3α(-k/3) = -k α3α^2β ={k+k[1 +3k]^0.5}/33α^2β ={1+[1 +3k]^0.5}k/3
Check: Let k = 1
3x^2+2x-1 = 0, (3x -1)(x + 1)
α =-1, β = 1/3
3α^2β ={1+[1 +3k]^0.5}k/3
3α^2β = 3(1)(1/3) = 1
{1+[1 +3k]^0.5}k/3 = (1+2)1/3 = 1

If the roots of the equation x^2+px-q=0 differ by2,show that p^2=4(1-q)α + β = -pα + (α + 2) = p2α +2 = p2(α +1) = p (α +1) = p/2 ----------------(1) α β = -qα (α + 2) = -qα ^2 + 2α = -qα ^2 + 2α + 1 = 1 -q(α + 1)^2 = 1 –q -------------(2)Substitute (1) into equation (2)[p/2]^2 =( 1 - q ) (p^2)/4 =( 1 - q ) p^2 = 4(1 - q)