Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 34

$y=2$

$Given,$ $2y=\sqrt (5y+6)$ Squaring both sides,we get: $4y^{2}=5y+6$ $4y^{2}-5y-6=0$ $(4y+3)(y-2)=0$ $y=2,y=\frac{-3}{4}$ We need to check by the putting the obtained solutions in the original equation. $y=2$ $L.H.S=2X2=4$ $R.H.S=\sqrt (5X2+6)=\sqrt 16=4$ $L.H.S=R.H.S$,hence it is a valid solution $y=\frac{-3}{4}$ $L.H.S=2X\frac{-3}{4}=\frac{-3}{2}$ $R.H.S=\sqrt (5X\frac{-3}{4}+6)=\sqrt \frac{24-15}{4}=\sqrt\frac{9}{4}=\frac{3}{2}$ $L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.