Question #a3836

#26-11x-2x^2=-2x^2-11x+26#

#-2x^2-11x+26# is not factorable.

The roots of #-2x^2-11x+26# are #x=-(11+sqrt(329))/4# and #x=(sqrt(329)-11)/4#.

#(x+(11+sqrt(329))/4)(x-(sqrt(329)-11)/4)=x^2+(11x)/2-13#

#:.-2(x^2+(11x)/2-13)=-2x^2-11x+26=-2(x+(11+sqrt(329))/4)(x-(sqrt(329)-11)/4)#

Factorise by completing the square;
#y=-2x^2-11x+26#
1) Convert to a monic quadratic function by dividing both sides of the equality by #-2#
#y/(-2)=(-2x^2-11x+26)/(-2)#

#-y/2=x^2+(11x)/2-13#

2) Add 13 to both sides
#-y/2+13=x^2+(11x)/2#

3) Add the square of half of the #x# term #11/2# (which is #11/4#) to both sides
#-y/2+13+(11/4)^2=x^2+(11x)/2+(11/4)^2#

4) Combine the like terms on the left of the equality
#-y/2+329/16=x^2+(11x)/2+(11/4)^2#

5) What we have been trying to do is create a perfect square on the right hand side of the equality. Now it is time to factorise that perfect square.

#-y/2+329/16=(x+11/4)^2#

6) Now we are going to move all of the stuff on the left of the equality back to the right (barring the #y# variable of course)

#-y/2=(x+11/4)^2-329/16#

#y=-2[(x+11/4)^2-329/16]#

7) Multiplying the #-2# into the brackets will give us our final answer

#y=-2(x+11/4)^2+329/8#

I hope that helps :)

Harold