# Please answer this question : the roots of ax^2+bx+c=0 where a is not equal to 0 and coefficients are real, are non real complex and a+c a)4a+c<2b b)4a+c>2b c)4a+c=2b d)none of these

1
by juttuc4
the roots of ax^2+bx+c=0 where a is not equal to 0 and coefficients are real, are non real complex and a+c<b then
a)4a+c<2b
b)4a+c>2b
c)4a+c=2b
d)none of these
It seems a+c<b is a wrong condition. In that case, b^2 -4ac >0 and the roots will be real. it is said that roots are imaginary. it should be a+c > b. It implies that b^2 - 4ac < 0.

2015-01-01T19:56:59+05:30

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Roots of  a x^2 + b x + c = 0   are    [ -b + - sqroot(b^2 - 4 a c) ] / 2

they are non real and complex.  Hence  b^2 < 4 a c.  --- equation 1

we know that a square of a real number is always non negative,
So  (4a - c)^2 >= 0
=>  16a^2 + c^2 >= 8 a c    -- eq 2

Now find (4 a + c)^2 = 16 a^2 + c^2 + 8 a c
>=  8 a c + 8 a c       --- by eq 2
> 4 * b^2         --- by eq 1

take square roots on both sides

Hence   4 a + c >  2 b

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