Solution of logx2+6x+8log2x2+2x+3(x2−2x)=0 is
a negative integer
-1
We must have
x2+6x+8>0,2x2+2x+3>0 and x2+6x+8≠1,2x2+2x+3≠1∴(x+4)(x+2)>0 and,(x+1/2)2+5/4>0⇒ xϵ(−∞,−4)∪(−2,∞) and x≠−3±√2 (1)and x2−2x>0 i.e.xϵ(∞,0)∪(2,∞)
The given equation can be written as
log2x2+2x+3(x2−2x)=1⇒ x2−2x=2x2+2x+3⇒ x2+4x+3=0⇒x=−1,−3x=−1 x ≠−3 (As it doesn't satisfy (1))