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The roots of ax 2 +bx+c=0, where a≠0. an coefficents are real,are non-real complex and a+c 1) 4a+c>2b 2) 4a+c 3)4a+c=2b 4) none

The roots of ax2+bx+c=0, where a≠0. an coefficents are real,are non-real complex and a+c

1) 4a+c>2b


2) 4a+c<2b


3)4a+c=2b


4) none

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) = ax^2 + bx + c
The roots of the equation f(x) = 0 are non-real complex.
So f(x) will not touch the real axis. Either it is above or below the real axis.
Given the condition
a + c < b
which can be interpreted as
f(-1) = a-b+c
f(-1) < 0
a -b + c < 0
a + c < b
So f(x) is -ve for x = -1. So it will be -ve for all x.
f(-2) = 4a-2b+c
f(-2) < 0
4a – 2b + c < 0
4a + c < 2b

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