Answer
$a=\frac{1}{4}$.
Work Step by Step
Given:
$\sqrt {1-3a}=2a$
Squaring both sides, we have
$1-3a=4a^{2}$
Adding $3a$ and $-1$ to both sides, we get
$0=4a^{2}+3a-1$
Factoring, we get
$(4a-1)(a+1)=0$
Using the Zero-product property, we obtain
$4a-1=0$ or $a+1=0$.
Solving for $a$, we get
$a=\frac{1}{4}$ or $a=-1$
Let's check the results:
When $a=\frac{1}{4}$,
$\sqrt {1-3(\frac{1}{4})}=\frac{1}{2}=2\left(\frac{1}{4}\right)\checkmark$
When $a=-1$,
$\sqrt {1-3(-1)}=\sqrt {4}=2\ne2(-1)$
$a=-1$ does not satisfy the original equation and is an extraneous solution. The only solution is $a=\frac{1}{4}$.
