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ax^2+bx=6=0, has no two real and distinct roots then find the value of 3a+b.



19-Dec-2015 4:33 PM
~6

Answers (1)


2
(I have seen this kind of question, and they actually ask the minimum value of 3a+b.... )

QE: ax+ bx + 6 = 0

Since the Quadratic Equation has no distinct roots, so b2 = (4)(a)(6)

Or, 3a = \frac{b^{2}}{8}

3a + b= \frac{b^{2}}{8} + b = \frac{b^{2}+8b}{8} = \frac{(b+4)^{2}-16}{8} = \frac{(b+4)^{2}}{8} - 2

 

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



19-12-2015 17:08
~32

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