F4 Maths (equation)

(square root(5x-16))=2-(square root(3x+4))
(square root(5x-16))^2=(2-(square root(3x+4)))^2
5x-16=4-4square root(3x+4)+3x+4
2x-24=-4square root(3x+4)
x-12=-2square root(3x+4)
(x-12)^2=(-2square root(3x+4))^2
x^2-24x+144=4(3x+4)
x^2-24x+144=12x+16
x^2-36x+128=0
x=4 or 32
sub x=4 into (square root(5x-16))=2-(square root(3x+4))
L.H.S.=(square root(5(4)-16))
=2
R.H.S.=2-(square root(3(4)+4))
=-2
L.H.S.=/=R.H.S.
sub x=32 into (square root(5x-16))=2-(square root(3x+4))
L.H.S.=(square root(5(32)-16)
=12
R.H.S.=2-(square root(3(32)+4))
=-8
L.H.S.=/=R.H.S.
the equation has no solutions.

√(5x-16)=2-√(3x+4)
range of values of x:
5x-16≥0 and 3x+4≥0
⇒x≥16/5 and x≥-4/3
⇒x≥16/5
Now √(5x-16)≥0
but 2-√(3x+4)≤2-√(16/5+4)<0
∴√(5x-16)≠2-√(3x+4) for all x≥16/5
the equation has no solution.