1
Question
If α and β are zeros of x2−3x−2 then find quadratic polynomial whose zeros are
1)12α+β and 12β+α
2)α−1α+1 and β−1β+1
1)12α+β and 12β+α
2)α−1α+1 and β−1β+1
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Solution
(1)α & β are zeroes if x2−3x−2.
∴α+β=3αβ=−2.
Nav zeroes,(N roots) 12α+β&12β+α
sum of roots = 12α+β+12β+α
=32(α+β)=32(3)
Product of roots = (α2+β)(β2+α)
=αβ4+β22+α22+αβ
=−24−4+12(α2+β2)=−2−164+12{(α+β)2−2αβ}=−14+12{32−2(−2)}=−184+12(9+4)=−184+132=−18+264=84=2.
∴ Required quadartic equation is
x2−(94)x+2=4x2−9x+8=0
(2)Now roots :α−1α+1&β−1β+1
sum of roots =α−1α+1=(α−1)(β+1)+(β−1)(α+1)(α+1)(β+1)
=αβ−β+α−1+αβ−α+β−1(α+1)(β+1)=2(αβ−1)αβ+β+α+1==2{(−2)−1}−2+3+1=−62=−3
Product of roots=(α−1)(β−1)(α+1)(β+1)=αβ−β−α+12=−2−3+12
=−42=−2
∴ Quadratic equation x2−(−3)x−2=0
x2+3x−2=0
∴α+β=3αβ=−2.
Nav zeroes,(N roots) 12α+β&12β+α
sum of roots = 12α+β+12β+α
=32(α+β)=32(3)
Product of roots = (α2+β)(β2+α)
=αβ4+β22+α22+αβ
=−24−4+12(α2+β2)=−2−164+12{(α+β)2−2αβ}=−14+12{32−2(−2)}=−184+12(9+4)=−184+132=−18+264=84=2.
∴ Required quadartic equation is
x2−(94)x+2=4x2−9x+8=0
(2)Now roots :α−1α+1&β−1β+1
sum of roots =α−1α+1=(α−1)(β+1)+(β−1)(α+1)(α+1)(β+1)
=αβ−β+α−1+αβ−α+β−1(α+1)(β+1)=2(αβ−1)αβ+β+α+1==2{(−2)−1}−2+3+1=−62=−3
Product of roots=(α−1)(β−1)(α+1)(β+1)=αβ−β−α+12=−2−3+12
=−42=−2
∴ Quadratic equation x2−(−3)x−2=0
x2+3x−2=0
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