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Question

If the equation ax2+bx+6=0 does not have two distinct real roots, then the least value of 3a+b

A
3
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B
-3
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C
2
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D
-2
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Solution

The correct option is D -2


Solution : f(x)=ax2+bx+6=0
f(0) = 6 positive
f(x)=ax2+bx+60 xR
f(3)=9a+3b+60
3(3a+b)6
3a+b2

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