If alpha and beta are zeroes of the polynomial ax2+bx+c, find a quadratic polynomial whose zeroes sre 1/a*alpha +b and 1/a*beta+b
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Step-by-step explanation:
ax
2
+bx+c=0
α+β=
a
−b
αβ=
a
c
(i) equation whose roots are
α
2
1
,
β
2
1
Sum of roots =
α
2
1
+
β
2
1
=
(αβ)
2
α
2
+β
2
=
(αβ)
2
(α+β)
2
−2αβ)
=
a
2
c
2
a
2
b
2
−
a
2c
=
c
2
b
2
−2ac
\
product =
α
2
1
β
2
1
=
c
2
a
2
∴ equation is
c
2
x
2
+(2ac−b
2
)x+a
2
=0
(ii)
aα+β
1
,
aβ+α
1
Sum ⇒
a
2
αβ+α
2
a+aβ
2
+αβ
aβ+aα+β
⇒
(a
2
+1)(αβ)+a(αβ)
(α+β)(1+a)
⇒
(a
2
+1)
a
c
+a(
a
2
b
2
−
a
2c
)
(1+a)(
a
b
)
⇒
(a
2
+1)c+a(b
2
−2ac)
−(1+a)b
⇒
c(1+a
2
)+ab
2
−(1+a)b
Product
c(1−a
2
)+ab
2
a
∴(c(1−a
2
)+ab
2
)x
2
+(1+a)bx+a=0
(iii) α+
β
1
,β+
α
1
Sum ⇒
β
αβ+1
+
α
αβ+1
⇒
αβ
α
2
β(αβ)+(α
2
+β
⇒
a
−b
+
c
−b
−b[
ac
a+c
]
Product
αβ
(αβ+1)
2
⇒
a
2
a
c
(c+a)
2
⇒(α+β)+
αβ
αβ
[
a+c
ac
]x
2
+bx+(a+c)=0
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