Sol.
Q.
If and β are the zeros of the quadratic polynomial f(x) = x2 – px + q,
Prove that
+
β2
2
2
β2
=
+ 2
–
q2
p4
q
4p4
f(x) = x2 – px + q
Q
and β are the zeros of f(x)
∴
+
β
=
a
–b
=
=
p
1
–
β
=
a
c
=
1
q
=
q
2 β2
(2)2
L.H.S. =
β2
2
+
2
β2
=
=
[(2 + β2)2
–
22β2]
2β2
( + β)2 – 2β
=
(β)2
–
2
+
(β2)2
🗹
and
Here
a
=
1,
b
=
–p,
c
=
q
1
2 + β2
{ }2
[
]
2β2
2
–
∴
(2)2
+
(β2)2
=
(
2
+
β2
)2
–
(β)2
2β2
(–p)
Sol.
f(x) = x2 – px + q
+
β
=
p
and
β
=
q
[
=
–
(q)2
2
(p2)2
=
+
(p)2
–
2
q
(q)2 }
–
(q)2
[
2
{
–
(p)2
(2q)
(2q)2
2(q)2
( + β)2 – 2β
(β)2
–
2
{ }2
[
]
(β)2
L.H.S. =
p4
=
+
4q2
–
(q)2
–
4p2q
2q2
2
p4 – 4p2q
=
+
q2
2q2
q2
2q2
q2
p4
–
q2
4p2q
+
=
2
q2
p4
–
q
4p2
+
=
=
R.H.S.
β2
2
+
2
β2
∴
=
q2
p4
+
q
4p4
+
2
Q.
If and β are the zeros of the quadratic polynomial f(x) = x2 – px + q,
Prove that
+
β2
2
2
β2
=
+ 2
–
q2
p4
q
4p4