I2.1 If the product of the real roots of the equation $x^{2}+9 x+13=2 \sqrt{x^{2}}+9 x+21$ is $P$, find the value of $P$.
Let $y=x^{2}+9 x$, then the equation becomes $y+13=2 \sqrt{y+21}$
$$
\begin{array}{l}
(y+13)^{2}=4 y+84 \\
y^{2}+26 y+169-4 y-84=0 \\
y^{2}+22 y+85=0 \\
(y+17)(y+5)=0 \\
y=-17 \text { or } y=-5
\end{array}
$$

Check put $y=-17$ into the original equation: $-17+13=2 \sqrt{-17+21}$
LHS $0$, rejected
Put $y=-5$ into the original equation: LHS $=-5+13=2 \sqrt{-5+21}=$ RHS, accepted
$$
\begin{array}{l}
x^{2}+9 x=-5 \\
x^{2}+9 x+5=0
\end{array}
$$

Product of real roots $=5$
Method 2
Let $y=\sqrt{x^{2}+9 x+21} \geq 0$
Then the equation becomes $y^{2}-8=2 y \Rightarrow y^{2}-2 y-8=0$
$$
\begin{array}{l}
(y-4)(y+2)=0 \Rightarrow y=4 \text { or }-2 \text { (rejected) } \\
\Rightarrow x^{2}+9 x+21=16 \\
x^{2}+9 x+5=0 \\
\Delta=9^{2}-4(5)>0
\end{array}
$$

Product of real roots $=5$