JEE Quadratic#7

Algebra Level 4

If the equation 0 = a x 2 + 2 b x − 3 c 0=ax^{2} +2bx-3c has non real roots and 3 c 4 < ( a + b ) \frac{3c}{4}< (a+b) ; then c is always

positive cannot be determined negative zero

Ninad Akolekar by Ninad Akolekar , 23, India

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2 solutions

Harry Potter
Apr 25, 2015

f(2)>0 as putting x=2,we get 4(a+b)-3c and it is given that 3c<4(a+b).now put x=0,so -3c should be greater than 0.therefore c is negative

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Himanshu Aswani
Mar 25, 2015

Put x=2 we get 4a+4b-3c .As 3c<4(a+b) therefore function is always of same sign as f(2) ;where f(x)=ax^2+2b*x-3c ;f(2)>0;put x=0 we get -3c which being greater than zero tells us c is negative.

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