If a,b,c are roots of the equation (x-2)(x^2 +6x-11)= 0. what is (a+b+c)

Question

sriram · Answer

23

anonymous · Answer

i guess you have a choice. for one thing  you can find the roots and  just add them  probably the easiest way

anonymous · Answer

it cant be 23

anonymous · Answer

it isn't 
it is -4

sriram · Answer

coz in ax^3+bx^2+cx+d
sum of roots=-b/a

anonymous · Answer

ya it is -4

anonymous · Answer

wolframmed??

sriram · Answer

oh sorry calculation errors !!!!

anonymous · Answer

wow site just went nuts
roots are 2 and 
$$-3\pm2\sqrt{5}$$

anonymous · Answer

but sriram is right it is 
$$-\frac{a_2}{a_3}=-\frac{4}{1}=-4$$

anonymous · Answer

that is why i was trying to say "you have a choice" but suddenly this site went haywire.
you can multiply out as well and get 
$$x^3+4 x^2-23 x+22 = 0$$

anonymous · Answer

kk
thanks

anonymous · Answer

still flashing wow
yw

phi · Answer

Another way to do this is recognize one root is +2 from (x-2)
the quadratic can be factored, but we know it will be 2 complex conjugate roots, so we just need the -b/2a twice, or -b/2 for the sum of the other 2 roots.