Hi Hansel,
It is mentioned in the question that the value of f(x) is always greater than or equal to zero.
Thus, the function will have only one root. This is because if f(x) has two roots (a,b where a is greater than b), then in the region a>x>b, the value of f(x) will be negative. So, f(x) will only have one root.
This is the mistake that you made in your solution.
Hope this helps. Please let us know if you face any issues.
Regards

Hi Pranjal,
It is given that $$f\left(x\right)\ge\ 0$$, which implies $$D\le\ 0$$. We know that if D<0, then the roots are not real. But in this case, we are getting a root, which is real, which implies D=0.
We know that if D=0, then the two roots are the same. In the case one root is 2, then another one is also 2.
Therefore, we can write f(x) as $$a\left(x-2\right)^2$$. ( since if we put x=2, then f(x) will be zero).
Hope this helps!

Hi,
Since there are three unknowns and you are getting only 2 equations, you will not be able to proceed further.
We need to use additional information given in the question. f(x) is greater than or equal to zero, which will allow us to write the quadratic equation in special form.
$$f\left(x\right)\ge\ 0\ =>\ D\le\ 0$$, and it is given that x =2 is a root of the equation, which implies D must be 0 ( if D is negative, then the roots are complex).
D = 0 => both the roots are the same, which implies both roots are 2.
Hence, we can write the equation in a special form, which is $$f\left(x\right)=a\left(x-2\right)^2$$.
Now you can proceed further since the number of unknowns is less compared to the previous scenario.
I hope this helps!

Hi Manisha,
We represented the equation in the form of $$a\left(x\ -\ b\right)^2\ +\ c$$, and the value of c here is not the same as c in the quadratic equation $$ax^2\ +\ bx\ +\ c$$ because the equation $$a\left(x-2\right)^2$$ has a c in the expansion of the terms. The expression $$a\left(x-2\right)^2\ =\ ax^2\ -\ 4ax\ +\ 4a$$ in which 4a is the value of c.
Hope this helps!

why take different quadratic eq please clarify because we 99% use the general quadratic eq not this if we use that how can we say constant is zero?

Hi Manisha,
Since there are three unknowns and you are getting only 2 equations, you will not be able to proceed further.
We need to use additional information given in the question. f(x) is greater than or equal to zero, which will allow us to write the quadratic equation in special form.
$$f\left(x\right)\ge\ 0\ =>\ D\le\ 0$$, and it is given that x =2 is a root of the equation, which implies D must be 0 ( if D is negative, then the roots are complex). D = 0 => both the roots are the same, which implies both roots are 2. Hence, we can write the equation in a special form, which is

$$f\left(x\right)=a\left(x-2\right)^2$$.
Now you can proceed further since the number of unknowns is less compared to the previous scenario.
I hope this helps!