Extinction chance when each cell has up to three offspring

A dish starts with a single bacterium. Each minute every bacterium, independently of the others, does one of four things with probability $\tfrac14$ each: dies, stays as it is, divides into two, or divides into three. The probability that the colony eventually dies out can be written as $\sqrt{a}-b$ with integers $a,b$. Find $a+b$.

$3$. Conditioning on the first minute, the extinction probability $x$ solves $x=\tfrac14(1+x+x^2+x^3)$, i.e. $(x-1)(x^2+2x-1)=0$. Since the mean offspring count is $1.5>1$, extinction is the smallest root in $[0,1]$, namely $x=\sqrt2-1$, so $a+b=2+1=3$.