If a is real, find the real values of
р which makes 4x²+px+1 always positive
Answers
Step-by-step explanation:
Explanation:
Given:
4
x
2
+
p
x
+
1
Note that this is in standard form:
a
x
2
+
b
x
+
c
with
a
=
4
,
b
=
p
and
c
=
1
It has discriminant
Δ
given by the formula:
Δ
=
b
2
−
4
a
c
=
p
2
−
4
(
4
)
(
1
)
=
p
2
−
16
=
(
p
−
4
)
(
p
+
4
)
Hence
Δ
<
0
whenever
−
4
<
p
<
4
When
Δ
<
0
the quadratic has no real zeros and hence does not intersect the
x
axis.
In addition note that as it is a polynomial it is defined for all values of
x
and continuous everywhere. So if it does not intersect the
x
axis, then it lies wholly on one side of the axis or the other.
Note that since the leading coefficient
4
is positive, the value of the quadratic will always be positive for large values of
x
.
So in order that it is positive for all values of
x
, we just require
−
4
<
p
<
4
.
Step-by-step explanation:
Given:
4
x
2
+
p
x
+
1
Note that this is in standard form:
a
x
2
+
b
x
+
c
with
a
=
4
,
b
=
p
and
c
=
1
It has discriminant
Δ
given by the formula:
Δ
=
b
2
−
4
a
c
=
p
2
−
4
(
4
)
(
1
)
=
p
2
−
16
=
(
p
−
4
)
(
p
+
4
)
Hence
Δ
<
0
whenever
−
4
<
p
<
4
When
Δ
<
0
the quadratic has no real zeros and hence does not intersect the
x
axis.
In addition note that as it is a polynomial it is defined for all values of
x
and continuous everywhere. So if it does not intersect the
x
axis, then it lies wholly on one side of the axis or the other.
Note that since the leading coefficient
4
is positive, the value of the quadratic will always be positive for large values of
x
.
So in order that it is positive for all values of
x
, we just require
−
4
<
p
<
4
.