Given that, $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$.
Let $f(x)=(x-a) g(x)$
On differentiating w.r.t. $x$, we get
$\Rightarrow f^{\prime}(x)=2(x-a) g(x)+(x-a)^2 g^{\prime}(x)$
Again, differentiating w.r.t $x$, we get
$\begin{aligned}
& \text { Now, } \begin{aligned}
& f^{\prime \prime}(x)=2 g(x)+2(x-a) g^{\prime}(x)+2(x-a) g^{\prime}(x) \\
&+(x-a)^2 g^{\prime \prime}(x)
\end{aligned} \\
& =2 g(x)+4(x-a) g^{\prime}(x)+(x-a)^2 g^{\prime \prime}(x)
\end{aligned}$
At $x=a$,
$\begin{aligned}
& \Rightarrow \quad f^{\prime}(a)=2(a-a) g(a)+(a-a)^2 g^{\prime}(a)=0 \\
& \Rightarrow \quad f^{\prime \prime}(a)=2 g(a)+4(a-a) g^{\prime}(a)+(a-a)^2 g^{\prime \prime}(a) \\
& \quad=2 g(a)
\end{aligned}$
Hence, $f(a)=f^{\prime}(a)=0, f^{\prime \prime}(a) \neq 0$