A) two distinct points whose coordinates are always rational numbers
B) no point
C) exactly two distinct points
D) exactly one point
Correct Answer: D
Solution :
Given \[a{{x}^{2}}+bx=c=0\]\[\Rightarrow \]\[a{{x}^{2}}=-bx-c\] \[y=4a{{x}^{2}}+3bx+2c\] \[=4[-bx-c]+3bx+2c\] \[=-4bx-4c+3bx+2c\] \[=-bx-2c\] Since, this curve intersects x-axis . \[\therefore \]put y = 0, we get\[-bx-2c=0\Rightarrow -bx=2c\]\[\Rightarrow \]\[x=\frac{-2c}{b}\] Thus, given curve intersects x-axis at exactly one point.
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