One root of the quadratic equation $2x^2+(k+3)x+2k=0$ is reciprocal of the other. Find the roots.

The correct option is -1 and -1.

We are given the quadratic equation:
$2x^2+(k+3)x+2k=0$

A standard quadratic equation is expressed in the form:
$a{x}^2+bx+c=0$

By comparing the given equation with the standard quadratic equation, we identify the coefficients as:

$a=2$

$b=k+3$

$c=2k$

Let the roots of the quadratic equation be $\alpha$ and its reciprocal $\frac{1}{\alpha }$.

The formula for the product of the roots of a quadratic equation is:
$\text{Product of roots}=\frac{c}{a}$

Substituting the roots and the values of $a$ and $c$:
$\alpha \cdot \frac{1}{\alpha }=\frac{2k}{2}$

$1=k$

Thus, we find that $k=1$.

Now, substitute $k=1$ back into the original quadratic equation:
$2x^2+(1+3)x+2\left(1\right)=0$

$2x^2+4x+2=0$

Divide the entire equation by 2:
$x^2+2x+1=0$

Express the left side as a perfect square:
${(x+1)}^2=0$

Solving for $x$:
$x=-1,-1$

Since the reciprocal of -1 is $\frac{1}{-1}=-1$, the condition that one root is the reciprocal of the other is satisfied.

Therefore, the roots of the equation are -1 and -1.