The solution of the equation p beta 2 + 1 is equal to 3 is
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Let α and β be the roots of equation px
2
+qx+r=0,p
=0. If p,q,r are in A. P. and
α
1
+
β
1
=4, then the value of ∣α−β∣
Hard
Solution
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Correct option is A)
GIven α,β are the roots of the equation px
2
+qx+r=0, p
=0
⇒α+β=−
p
q
and ⇒αβ=
p
r
⇒∣α−β∣
2
=α
2
+β
2
−2αβ
=(α+β)
2
−4αβ
=
p
2
q
2
−4
p
r
=(
p
q
)
2
−4(
p
r
) ...(1)
Also,
p
2q
=1+
p
r
...(2)
Given that,
α
1
+
β
1
=4
⇒
αβ
α+β
=4
⇒−
p
q
⋅
r
p
=4
⇒
r
q
=−4
Given p,q,r are in A.P.
⇒2q=p+r
⇒
r
2q
=
r
p
+1
⇒
r
p
=
r
2q
−1
⇒
r
p
=2(−4)−1 ...(∵
r
q
=−4)
⇒
r
p
=−9
⇒
p
r
=−
9
1
Substitute tha value of
p
r
in equation (2), we get,
p
2q
=1−
9
1
⇒
p
q
=
2
1
(1−
9
1
)
⇒
p
q
=
2
1
(
9
9−1
)
⇒
p
q
=
2
1
⋅
9
8
⇒
p
q
=
9
4
Now, substitute the values of
p
q
and
p
r
in the equation (1), we get
⇒∣α−β∣
2
=(
9
4
)
2
−4(−
9
1
)
⇒∣α−β∣
2
=(
9
4
)
2
+
9
4
⇒∣α−β∣
2
=
9
4
(
9
13
)
⇒∣α−β∣
2
=
81
52
⇒∣α−β∣=
9
2
13