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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.

10 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, Kushagra Madhukar
one year ago
Dear student,

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