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Byju's Answer
Standard X
Mathematics
Discriminant
If the equati...
Question
If the equation
a
x
2
+
b
x
+
c
=
0
has non-real roots,prove that
1
+
c
a
+
b
a
>
0
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Solution
The equation
a
x
2
+
b
x
+
c
=
0
is non-real
⇒
b
2
−
4
a
c
<
0
⇒
b
2
<
4
a
c
⇒
|
b
|
<
2
√
a
c
⇒
b
<
a
+
c
⇒
a
+
c
−
b
>
0
⇒
a
+
b
+
c
>
0
As A.M
>
G.M
⇒
a
+
c
2
>
√
a
c
⇒
a
+
c
>
2
√
a
c
⇒
a
+
c
2
>
√
a
c
⇒
−
b
<
√
a
c
⇒
a
+
b
+
c
>
0
⇒
a
a
+
b
a
+
c
a
>
0
by dividing by
a
⇒
1
+
b
a
+
c
a
>
0
Hence proved
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0
Similar questions
Q.
If a, b, c are real numbers such that ac
≠
0, then show that at least one of the equations
a
x
2
+ bx + c = 0 and -
a
x
2
+ bx + c = 0 has real roots.
Q.
For the equation
a
x
2
+
b
x
+
c
=
0
,
a
,
b
and
c
are real,
Statement 1: If the equation
a
x
2
+
b
x
+
c
=
0
,
0
<
a
<
b
<
c
, has non-real complex roots
z
1
and
z
2
, then
|
z
1
|
>
1
,
|
z
2
|
>
1
.
Statement 2: Complex roots always occur in conjugate pairs.
Q.
Assertion :If the equation
a
x
2
+
b
x
+
c
=
0
,
0
<
a
<
b
<
c
,
has non real complex roots
z
1
and
z
2
, then
|
z
1
|
>
1
,
|
z
2
|
>
1
. Reason: Complex roots always occur in conjugate pairs.
Q.
A quadratic equation
a
x
2
+
b
x
+
c
=
0
has no real roots if:
Q.
If
a
,
b
,
c
∈
R
and the quadratic equation
a
x
2
+
b
x
+
c
=
0
has no real roots, then
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