A) \[y={{x}^{2}},3x-2y-3=0\]
B) \[y=4{{x}^{2}},6x-2y-3=0\]
C) \[y=3{{x}^{2}},6x-y-3=0\]
D) \[y=2{{x}^{2}},6x+2y-3=0\]
Correct Answer: D
Solution :
(d): Take 6x ? 3 on other side: \[4{{x}^{2}}+6x-3=0\]which can be written as \[2{{x}^{2}}=-\left( 3x-\frac{3}{2} \right)\] \[~\therefore y=2{{x}^{2}}\]and\[y=-\left( 3x-\frac{3}{2} \right)\]will intersect to give roots.
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