If the sum of the roots of the quadratic
equation ax? + bx + c = 0 is equal to
the sum of the squares of their
a b c
reciprocals, then a/c ,b/a,c/a are in
Answers
Answer:
HP
ax
2
+bx+c=0
The sum of roots are α+β=
a
−b
=
α
2
1
+
β
2
1
=
(αβ)
2
α
2
+β
2
=
(αβ)
2
(α+β)
2
−2(αβ)
The product of the roots are αβ=
a
c
Substituting α+β=
a
−b
and αβ=
a
c
, we get
a
−b
=
(
a
c
)
2
(
a
−b
)
2
−2(
a
c
)
a
−b
=
(
a
c
)
2
(
a
b
)
2
−2(
a
c
)
=
c
2
b
2
−2ac
Cross multiplying, −bc
2
=ab
2
−2a
2
c
Rearranging the terms,
ab
2
+bc
2
=2a
2
c
Dividing throughout by abc,
c
b
+
a
c
=
b
2a
This is true if
c
b
,
b
a
,
a
c
are in AP.
Hence their reciprocal,
b
c
,
a
b
,
c
a
are in HP.
Step-by-step explanation:
I hope you understand this answer
H.P.
ax
2
+bx+c=0
Let α,β be the roots of the equation
⇒α+β=
a
−b
,αβ=
a
c
………..(1)
⇒α+β=
α
2
1
+
β
2
1
[Given]
⇒α+β=
α
2
β
2
α
2
+β
2
⇒α+β=
α
2
β
2
(α+β)
2
−2αβ
………….(2) substituting the values of equation (1) in equation (2)
⇒
a
−b
=
a
2
c
2
a
2
b
2
−
a
2c
⇒
a
−b
=
c
2
b
2
−2ac
⇒−bc
2
=ab
2
−2a
2
c
⇒2a
2
c=ab
2
+bc
2
Then ab
2
,a
2
c,bc
2
are in A.P
Dividing by abc, we get:-
c
b
,
b
a
,
a
c
are in AP
⇒
a
c
,
b
a
,
c
b
are in AP
⇒
c
a
,
a
b
,
b
c
are in HP.