Read the following quadric equation carefully and answer the questions given below.

Equation 1: $a^2-7a+d=0$

Equation 2: $b^2-4b+(d-9)=0$

Note: Roots of the equation 1 are ‘x’ and ‘y’ and the roots of equation 2 are ‘x’ and ‘y – x’ .

If we multiply equation 2 by 3^z and add 1 in the smallest root of the equation newly formed, then find the resultant number?

The correct answer is 2.

Let's solve the problem step-by-step:

Step 1: Understand the roots of Equation 1
Equation 1 is given as:
$a^2-7a+d=0$
The roots of Equation 1 are given as $x$ and $y$.
Using the relationship between the coefficients and the roots of a quadratic equation:
Sum of the roots:
$x+y=7$ (Equation 3)
Product of the roots:
$x·y=d$ (Equation 4)

Step 2: Understand the roots of Equation 2
Equation 2 is given as:
$b^2-4b+(d-9)=0$
The roots of Equation 2 are given as $x$ and $y-x$.
Using the relationship between the coefficients and the roots of Equation 2:
Sum of the roots:
$x+(y-x)=4$
Simplifying this, we get:
$y=4$

Step 3: Solve for the variables
Substitute the value of $y=4$ into Equation 3:
$x+4=7$
$x=3$
Now, find the value of $d$ using Equation 4:
$d=x·y=3·4=12$

Step 4: Find the roots of Equation 2
The roots of Equation 2 are:
First root: $x=3$
Second root: $y-x=4-3=1$
Thus, the roots of Equation 2 are 3 and 1, with the smallest root being 1.

Step 5: Apply the transformation to the equation
If we multiply Equation 2 by any non-zero constant $3^z$, the roots of the equation do not change.
Therefore, the roots of the newly formed equation remain 3 and 1, and the smallest root is still 1.
Adding 1 to the smallest root gives:
$1+1=2$