Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Let graph of f(x)=a x2+b x+c is as shown in figure. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/mathematics/87aad52f1986d77fc3eb0fe66ec70937-.png /> |(b2/4 a)|-|c| is
Q. Let graph of
f
(
x
)
=
a
x
2
+
b
x
+
c
is as shown in figure.
∣
∣
4
a
b
2
∣
∣
−
∣
c
∣
is
1285
195
Complex Numbers and Quadratic Equations
Report Error
A
Positive
B
zero
C
negative
D
can't say
Solution:
∣
∣
4
a
b
2
∣
∣
−
∣
c
∣
=
−
4
a
b
2
+
c
=
4
a
−
[
b
2
−
4
a
c
]
>
0