Here,
#8x^2=-11x-7#
#=>8x^2+11x+7=0#
#=>8x^2+11x=-7#
Let , #color(violet)(kinRR # be the #3^(rd) term # to complet square.
#i.e. 8x^2+11x+color(violet)(k)=color(violet)(k)-7...to(1)#
In the #LHS# we have ,
#color(blue)(diamond 1^(st)term=8x^2#
#color(blue)(diamond2^(nd)term=11x#
#color(blue)(diamond3^(rd)term)=color(violet)(k#
We have formula for #3^(rd)term :#
#color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))...to(A)#
#=>color(violet)(k)=(11x)^2/(4xx8x^2)=(121x^2)/(32x^2)#
#=>color(violet)(k=121/32#
From #(1)#,we get
#8x^2+11x+color(violet)(121/32)=color(violet)(121/32)-7=-103/32#
#=>(2sqrt2x)^2+2(2sqrt2x)(11/(4sqrt2))+(11/(4sqrt2))^2=103/32*i^2#
#=>(2sqrt2x+11/(4sqrt2))^2=(sqrt103/(4sqrt2)*i)^2#
#=>2sqrt2x+11/(4sqrt2)=+-sqrt103/(4sqrt2)*i#
#=>2sqrt2x=-11/(4sqrt2)+-sqrt103/(4sqrt2)*i#
#=>x=-11/16+-sqrt103/16*i#
...................................................................................................
Note :
Formula #(A) :color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))# can be use
to find THIRD TERM for any eqn. without any doubt .
WHY ??? #to#Please see below.
#diamond if , a^2+2ab+k=0# ,then [use #(A)#]
#k=(2ab)^2/(4xxa^2)=(4a^2b^2)/(4a^2)=b^2#
#=>a^2+2ab+b^2=(a+b)^2#
