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4-2.Quadratic Equations and Inequations
hard
If $a, b, c \in R$ and $1$ is a root of equation $ax^2 + bx + c = 0$, then the curve y $= 4ax^2 + 3bx+ 2c, a \ne 0$ intersect $x-$ axis at
A
two distinct points whose coordinates are always rational numbers
B
no point
C
exactly two distinct points
D
exactly one point
(AIEEE-2012)
Solution
Given $ax^{2}+bx+c=0$
$\Rightarrow ax^{2}=-bx-c$
Now, consider
$y=4ax^{2}+3 bx+2 c$
$=4[-b x-c]+3 b x+2 c$
$=-4 b x-4 c+3 b x+2 c$
$=-b x-2 c$
since, this curve intersects $x$ -axis
$\therefore$ put $y=0,$ we get
$-b x-2 c=0 \Rightarrow-b x=2 c$
$\Rightarrow x=\frac{-2 c}{b}$
Thus, given curve intersects $x$ -axis at exactly one point.
Standard 11
Mathematics