yeah wtf? how does that work?Originally posted by klaw
3. a^2+b^2+c^2=(a+b+c)³-3ab-3ac
=(a+b+c)²-3(ab+ac)
=1-3(-2)
=7
6(a^3+b^3+c^3)+6(a^2+b^2+c^2)-12(a+b+c)+9=0Yip said:Incorrect again klaw, but that one was quite tricky. FOr that one since a, b and c are roots, u substitute them in to get a system of 3 simulatenous equations:
6a^3+6a^2-12a+3=0
6b^3+6b^2-12b+3=0
6c^3+6c^2-12c+3=0
Adding the equations,
6(a^3+b^3+c^3)+6(a^2+b^2+c^2)-12(a+b+c)+9=0
6(a^3+b^3+c^3)+6(5)-12(-1)+9=0
6(a^3+b^3+c^3)=-51
therefore a^3+b^3+c^3=-51/6
At (-4,10),What is the equation of the tangent to the parabola f(x) = x² + x - 2 at the point where x = -4?
f(-4)=16-4-2=10insert-username said:What is the equation of the tangent to the parabola f(x) = x² + x - 2 at the point where x = -4?
x² = 12yinsert-username said:Tee hee. Errors all over the shop.
Find the focal length, the equation of the directrix, and the coordinates of the focus of the parabola with equation x² = 12y
Find the equation of the normal to the tangent to the parabola x² = 8y at the point where x = 4
I_F
ohh oops.. must coz i made this thread.. =Pklaw said:this thread is cursed... so many silly mistakes...