Answer
$252$
Work Step by Step
We are given the expression:
$\left(\sqrt x+\dfrac{3}{\sqrt x}\right)^{8}$
The term $T_{k+1}$ of the expansion of $(a+b)^n$ is:
$T_{k+1}=\binom{n}{k}a^{n-k}b^r$
We have:
$T_{k+1}=\binom{8}{k}(x^{1/2})^{8-k}(3x^{-1/2})^k=3^k\binom{8}{k}x^{4-0.5k-0.5k}=3^k\binom{8}{k}x^{4-k}$
Determine $k$ so that the term contains $x^2$:
$4-k=2$
$k=2$
Determine the coefficient of $x^2$:
$3^2\binom{8}{2}=9\cdot\dfrac{8!}{2!6!}=252$
