The correct option is D None of these.
6√xx+4−2√x+4x=11
For the square roots to exist
xx+4≥0⇒x<−4 or x≥0⋯(1)
x+4x≥0⇒x≤−4 or x>0⋯(2)
From equation (1) and (2),
x∈(−∞,−4)∪(0,∞)
Assuming √xx+4=y
The given equation can be written as,
6y−2y=11⇒6y2−11y−2=0
⇒(6y+1)(y−2)=0⇒y=−16, y=2⇒y=2 (∵y=√xx+4≥0)
Thus,
y=2⇒xx+4=y2=4⇒x=4x+16⇒x=−163
This is a acceptable value of x as x∈(−∞,−4)∪(0,∞)