- A$1$
- B$2$
- C$3$
- ✓$4$
So, $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = \frac{{{{(a - c)}^2}}}{{\left\{ {{{\left( {\frac{{a + c}}{2}} \right)}^2} - ac} \right\}}}$
$ = \frac{{{{(a - c)}^2}4}}{{[{a^2} + {c^2} + 2ac - 4ac]}} = \frac{{4{{(a - c)}^2}}}{{{{(a - c)}^2}}} = 4$.
Trick : Put $a = 1,\;b = 2,\;c = 3$,
then the required value is $\frac{4}{1} = 4$.
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Statement $-1$ : Equation of other roots is $x^2 + cx + ab = 0$
Statement $-2$ : $a + b + c = 0$
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is