# Please give answer as soon as possible........ Solve for x 2^|x+1|+2^|x|=6

Vikas TU
14149 Points
4 years ago
Hiii
There will be two cases for this question ,
In case 1 – Open x+1 and x as positive quantity  and solve for value of x,
In Case -2 – Open it as -(x+1) and -x , and solve for value of x ,
Take the intersection of both case- 1 and case -2 , that will be your solution ,
Hope this hep ,
If does not clear  ask me in the comment section, i will give solutions.
2081 Points
4 years ago
hello king. vikas’ answer above is a bit incomplete
case 1: when x belongs to (-inf, – 1]
the eqn becomes 2^-(x+1)+2^(-x)=6
write it as 2^(-x)+2*2^(-x)= 12
or 3*2^(-x)= 12
or 2^(-x)= 4
or x= – 2
case 2: when x belongs to (-1, 0]
the eqn becomes 2^(x+1)+2^(-x)=6
putting y= 2^x we get
2y^2+1= 6y
or 2y^2 – 6y+1=0
or y= (3 ± sqrt(7))/2
but x lies in (-1, 0], so 2^x=y must be lie in (½, 1]. obviously (3 + sqrt(7))/2 is greater than 1 while (3 – sqrt(7))/2 is less than ½. so both (3 – sqrt(7))/2 and (3+sqrt(7))/2 are rejected. so there is no solution in this interval.
case 3: when x belongs to (0, inf)
2^(x+1)+2^x= 6
2^x(2+1)= 6
or 2^x= 2
or x= 1
so the final soln is x= – 2, 1
kindly approve :)