Solve both equations and form a new equation in variable ‘z’ (reduce to lowest possible factor) using roots of equation 1 and 2 as per instructions given below.

1. $2-\frac{11}{x}+\frac{9}{x^2}=0$
2. ${(y-2)}^2=2\frac{1}{4}$

What will be new equation if roots of this are highest root of equation 1 and lowest root of equation 2.

To find the new equation in the variable $z$, we first need to solve both of the given equations and determine their roots.

Step 1: Solve Equation 1
The first equation is:
$2-\frac{11}{x}+\frac{9}{x^2}=0$
To solve this, we can multiply the entire equation by $x^2$ (assuming $x\ne 0$) to clear the denominators:
$2x^2-11x+9=0$
We can solve this quadratic equation by factoring. We look for two numbers that multiply to $2\times 9=18$ and add up to $-11$. These numbers are $-2$ and $-9$.
Splitting the middle term:
$2x^2-2x-9x+9=0$
$2x(x-1)-9(x-1)=0$
$(2x-9)(x-1)=0$
This gives the roots:
$x=\frac{9}{2}=4.5$ and $x=1$
Thus, the highest root of equation 1 is:
$z_1=\frac{9}{2}$

Step 2: Solve Equation 2
The second equation is:
${(y-2)}^2=2\frac{1}{4}$
First, convert the mixed fraction to an improper fraction:
$2\frac{1}{4}=\frac{2\times 4+1}{4}=\frac{9}{4}$
So the equation becomes:
${(y-2)}^2=\frac{9}{4}$
Taking the square root on both sides:
$y-2=\pm \frac{3}{2}$
This gives two cases for $y$:
Case 1: $y=2+\frac{3}{2}=\frac{7}{2}=3.5$
Case 2: $y=2-\frac{3}{2}=\frac{1}{2}=0.5$
Thus, the lowest root of equation 2 is:
$z_2=\frac{1}{2}$

Step 3: Form the new equation in variable 'z'
The roots of the new equation are $z_1=\frac{9}{2}$ and $z_2=\frac{1}{2}$.
Let us find the sum and the product of these roots:
Sum of roots $S=z_1+z_2=\frac{9}{2}+\frac{1}{2}=\frac{10}{2}=5$
Product of roots $P=z_1\times z_2=\frac{9}{2}\times \frac{1}{2}=\frac{9}{4}$
The standard form of a quadratic equation is:
$z^2-Sz+P=0$
Substituting $S$ and $P$:
$z^2-5z+\frac{9}{4}=0$
Multiplying the entire equation by 4 to reduce it to integer coefficients:
$4z^2-20z+9=0$

Therefore, the correct option is:
$4z^2-20z+9=0$