The length of the chord of the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{2}=1$ whose midpoint is $\left(1,\dfrac{1}{2}\right)$ is:

Given ellipse: $\frac{x^2}{4}+\frac{y^2}{2}=1$
Midpoint $(1,\frac{1}{2})$

Use midpoint (chord) formula: $T=S_1$
$\frac{xx_1}{4}+\frac{yy_1}{2}= \frac{x_1^2}{4}+\frac{y_1^2}{2}$

RHS:
$\frac{1}{4}+\frac{1/4}{2}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

So chord equation:
$\frac{x}{4}+\frac{y}{4}=\frac{3}{8}$
⇒ $2x+2y=3$ ⇒ $x+y=\frac{3}{2}$

Now find intersection with ellipse:
$y=\frac{3}{2}-x$

Substitute:
$\frac{x^2}{4}+\frac{(3/2-x)^2}{2}=1$

Multiply by 4:
$x^2+2\left(\frac{9}{4}-3x+x^2\right)=4$

$x^2+\frac{9}{2}-6x+2x^2=4$

$3x^2-6x+\frac{9}{2}-4=0$

$3x^2-6x+\frac{1}{2}=0$

$6x^2-12x+1=0$

Roots difference:
$|x_1-x_2|=\frac{\sqrt{144-24}}{6}$