Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2054 Upvotes
Verified Answer
The correct answer is:
exactly one point
exactly one point
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.
$$
\begin{aligned}
& \Rightarrow a x^2=-b x-c \\
& \text { Now, consider } \\
& y=4 a x^2+3 b x+2 c \\
& =4[-b x-c]+3 b x+2 c \\
& =-4 b x-4 c+3 b x+2 c \\
& =-b x-2 c
\end{aligned}
$$
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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.