Given quadratic equation is: $ax^{2}-bx+c=0$
Method 1.
Given it has two roots and both are $\beta$.
Using Shreedhara Acharya's formula we know,
$\beta =\frac{b- \sqrt{b^{2}-4ac}}{2a} \text{or}$
$\beta =\frac{b+ \sqrt{b^{2}-4ac}}{2a}$
So, $\beta +\beta =\frac{2b}{2a}=\frac{b}{a}$
$ \implies \beta =\frac{b}{2a}$
Now, $\beta *\beta =\frac{b-\sqrt{b^{2}-4ac}}{2a}*\frac{b+\sqrt{b^{2}-4ac}}{2a}$
$= \beta *\beta =\frac{b^{2}-(b^{2}-4ac)}{4a^{2}}=\frac{c}{a}$
$\beta ^{2}=\frac{c}{a}$
So, $\beta ^{3}=\beta *\beta ^{2}=\frac{b}{2a}*\frac{c}{a} \implies \beta ^{3}=\frac{bc}{2a^{2}}$
By this we can eliminate option A and B.
Option C is true.
Now option D,
$\beta ^{2}=\frac{c}{a}$ ……...(1)
$\beta ^{2}=\frac{b^{2}}{4a^{2}}$ ………...(2)
$\frac{c}{a}=\frac{b^{2}}{4a^{2}}$
or, $b^{2}=4ac$
Now we can say the $\beta^{2} \text{ value is either }\frac{b^2}{4a^2} \text{ or } \frac{c}{a}$.
Example:-
$4x^{2}-4x+1=0$ $(\text{here }a=4,b=4, c=1)$
Here roots($\beta$) are $\frac{1}{2},\frac{1}{2}$.
option A: $\frac{b}{a}=\frac{4}{4}=1\neq \beta$
option B: $a*c=4*1=4\neq \beta ^{2}$
option C: $\frac{bc}{2a^{2}}=\frac{4*1}{2*16}=\frac{1}{8}= \beta ^{3}$ (True)
option D: $4*a*c=4*4*1=16\neq \beta ^{2}$ (True for this case)
If we take the example
$\frac{1}{2}x^{2}-x+\frac{1}{2}=0$ $(\text{here }a=\frac{1}{2}, b=1,c=\frac{1}{2})$.
$\beta =1$
$\beta ^{2}=4*\frac{1}{2}*\frac{1}{2}=4ac$
so option D is wrong.
Correct answer is option c.