Given equation
√x+5+√x+21=√6x+40
For the square root to exist,
x+5≥0⇒x≥−5x+21≥0⇒x≥−216x+40≥0⇒x≥−406∴x≥−5
Squaring both sides,
x+5+x+21+2√(x+5)(x+21)=6x+40
⇒√(x+5)(x+21)=2x+7
Squaring both sides, we get
⇒(x+5)(x+21)=(2x+7)2⇒x2+26x+105=4x2+28x+49⇒3x2+2x−56=0⇒(x−4)(3x+14)=0
⇒x=4,−143
Checking validity for both the roots,
When x=4
⇒3+5=8→satisfy the given equationWhen x=−143⇒√13+√493=√12→Given equation is not satisfied
Hence, x=4 is the only root.