Answer with explanation:
It is given that vertex of Quadratic function has x coordinate equal to 3,and it has two distinct roots.
Suppose ,the vertex of Quadratic function ,which will be in the shape of parabola has coordinates ,(3,k).
→→Equation of Parabola ,having Vertex (3,k) ,can open in either of four directions that is along positive x-axis or negative x axis,or positive y-axis or negative y-axis.
So, any of the four equation are possible
1.→ y-b=(x-3)²
2. → y-c= - (x-3)²
3.→x-3=(y-m)²
4. → x-3= -(y-n)²
If you chose to find roots ,taking y as a independent variable,and x as a dependent variable, then you have to consider these two equations
3.→x-3=(y-m)²
4. → x-3= -(y-n)²
Such a two degree equation is
3. →x=(y-m)²+3
or,4. →x= -(y-n)²+3
→If you substitute ,the value of m by any real number in equation 3,it will not give real root.
So,this kind of Quadratic is not possible .
→→Now , in equation 4, if you will replace , n by any real number , and equate it with ,0 it will give two distinct real roots.
≡Now, Coming to Equation (1) and Equation (2)
1.→ y-b=(x-3)²
2. → y-c= - (x-3)²
here,if we consider , y as a dependent variable and x as an independent variable ,and then form Quadratic having x as a variable
1. y=(x-3)²+b
2. y= -(x-3)²+c
→In equation,1 if we limit the value of b from (-∞,0), and equate it with zero,there will be two real distinct roots.
⇒⇒In equation 2,if we limit the value of c from (0,∞),and equate it with zero,definitely there will be two distinct real roots.