GATE 2020 EC GA Q.9: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

1. The equation is quadratic, so $a\ne 0$. It has equal roots, which is $\beta$, so both of its roots are $\beta$.

2. Compare $a{x}^2-bx+c=0$ with the standard form $a{x}^2+px+q=0$. Here $p=-b$ and $q=c$.

3. Sum of the roots is $-\frac{p}{a}$: $$\beta +\beta =\frac{b}{a}\Rightarrow \beta =\frac{b}{2a}$$

4. Product of the roots is $\frac{q}{a}$: $$\beta \cdot \beta =\frac{c}{a}\Rightarrow {\beta }^2=\frac{c}{a}$$

5. Multiply the two results: $${\beta }^3=\beta \cdot {\beta }^2=\frac{b}{2a}\cdot \frac{c}{a}=\frac{bc}{2a^2}$$ This holds for every such equation, so (C) is always true.

6. (A) says $\beta =b/a$, but $\beta =\frac{b}{2a}$. The two agree only when $b=0$, so (A) is not true in general. (B) says ${\beta }^2=ac$, but ${\beta }^2=\frac{c}{a}$. These agree only when $c=0$ or $a^2=1$, so (B) is not true in general. For example, $2x^2-4x+2=0$, that is $2(x-1{)}^2=0$, has $\beta =1$, while $b/a=2$ and $ac=4$. (C) checks out here: $\frac{bc}{2a^2}=\frac{4\cdot 2}{2\cdot 4}=1={\beta }^3$.

7. Equal roots mean the discriminant is zero: $(-b{)}^2-4ac=0$, so $b^2=4ac$. That makes (D), $b^2\ne 4ac$, always false.

8. The only statement that must hold is (C).