Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 10

$1$

Add $1$ to both sides: $4\sqrt {x}-1+1=3+1$ $4\sqrt {x}=4$ Divide $4$ to both sides: $\dfrac{4\sqrt {x}}{4}=\dfrac{4}{4}$ $\sqrt {x}=1$ Square both sides. $\left(\sqrt {x}\right)^2=1^2$ $x=1$ Substitute into original problem to check for extraneous solutions: $4\sqrt {(1)}-1=3$ $4(1)-1=3$ $4-1=3$ $3=3$