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

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