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# If ax2+bx+6=0  does not have two distinct real roots,then the least value of 3a+b = ?Explain the answer.

## 2 Answers

9 years ago

Hi Pooja,

Since The QE has no distinct roots, so b^2 = 24a

Or, 3a = b^2 / 8

So 3a+b = b + b^2/8 =(8b+b^2)/8 = {(b+4)^2 - 16}/8 = (b+4)^2/8 - 2.

The minimum value of this will clearly be -2 -------------(when b = -4 and 3a = 2, or a=2/3)

That solves the Problem.

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras). Kushagra Madhukar
askIITians Faculty 629 Points
9 months ago
Dear student,
Please find the solution to your problem.

Since The QE has no distinct roots that means both of its roots are equal
Hence, D = 0
or, b2 – 4ac = 0
b2 = 24a
Or, 3a = b2 / 8
Now, 3a+b = b + b2/8
=(8b + b2)/8 = {(b+4)2 - 16}/8
= (b+4)2/8 – 2
The minimum value of any square term can be 0
Hence, The minimum value of 3a + b will clearly be – 2
(when b = -4 and 3a = 2, or a=2/3)

Hope this helps.
Thanks and regards,
Kushagra

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