If $a, b, c \in \mathrm{R}$ and 1 is a root of equation $a x^2+b x$ $+c=0$, then the curve $y=4 a x^2+3 b…
If $a, b, c \in \mathrm{R}$ and 1 is a root of equation $a x^2+b x$ $+c=0$, then the curve $y=4 a x^2+3 b x+2 c, a \neq 0$ intersect $x$-axis at
two distinct points whose coordinates are always rational numbers
no point
exactly two distinct points
exactly one point
Solution
Given $a x^2+b x+c=0$
$
\begin{aligned}
& \Rightarrow a x^2=-b x-c \\
& \text { Now, consider } \\
& y=4 a x^2+3 b x+2 c \\
& =4[-b x-c]+3 b x+2 c \\
& =-4 b x-4 c+3 b x+2 c \\
& =-b x-2 c
\end{aligned}
$
Since, this curve intersects $x$-axis
$\therefore$ put $y=0$, we get
$
\begin{aligned}
& -b x-2 c=0 \Rightarrow-b x=2 c \\
& \Rightarrow x=\frac{-2 c}{b}
\end{aligned}
$
Thus, given curve intersects $x$-axis at exactly one point