if alpha and beta are the zeros of the quadratic polynomial PX equal to X square + bx + c then find the value of1/alpha ²+1/beta²
Answers
Step-by-step explanation:
If α,β are roots of equation ax
2
+bx+c=0 then α+β=
a
−b
&αβ=
a
c
i) (1+α)(1+β)
=1+α+β+αβ
=1+(
a
−b
)+
a
c
=1+(
a
c−b
)
ii) α
3
β+αβ
3
αβ(α
2
+β
2
)
a
c
(
a
2
b
2
−
a
2c
)
a
3
c
(b
2
−2ac)
α+β=−
a
b
(α+β)
2
=b
2
/a
2
α
2
+β
2
+2αβ=b
2
/a
2
α
2
+β
2
=
a
2
b
2
−
a
2c
iii)
α
1
+
β
1
αβ
β
+
αβ
α
=
αβ
β+α
=
a
−b
×
c
a
=
c
−b
iv)
aα+b
1
+
aβ+b
1
α+β=
a
−b
=
aα−a(α+β)
1
+
aβ−a(α+β)
1
⇒b=−a(α+β)
=
aα−aα−aβ
1
+
aβ−aα−aβ
1
=−
a
1
(
α
1
+
β
1
)=
a
−1
(
c
−b
)=
ac
b
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Answer:
α,β are roots of equation ax
2+bx+c=0 then α+β= a−b &αβ=ac
i) (1+α)(1+β)
=1+α+β+αβ
=1+( a−b )+ ac
=1+( ac−b )
ii) α 2 β+αβ 3αβ(α 2 +β 2 )a c ( a 2b 2 − a2c )a 3c (b 2 −2ac)α+β=− ab
(α+β) 2 =b 2 /a
2α 2 +β 2 +2αβ=b 2/a 2α 2 +β 2 =a 2b 2 − a2c
iii)
α1 +β 1αβ + αβ α =αβ β+α= a−b × ca= c−b
iv) aα+b1 + aβ+b1α+β= a−b=
aα−a(α+β)1 + aβ−a(α+β)1
⇒b=−a(α+β)= aα−aα−aβ1 + aβ−aα−aβ1=− a1 ( a1 + β-1 )= a−1 ( c−b )= acb