If ααα and βββ are the zeros of polynomial x2x6,x2x6,x2x6, find a polynomial whose zeros are α2β2α2β2α2β2 and β2α2.β2α2.β2α2.


Answer:

k(x29736x+1)k(x29736x+1)k(x29736x+1)

Step by Step Explanation:
  1. On comparing the polynomial x2x6,x2x6,x2x6, with the standard form ax2+bx+c=0ax2+bx+c=0ax2+bx+c=0, we get:
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  2. Since, sum of zeros α+β=ba=11=1α+β=ba=11=1
    Product of zeros αβ=ca=61=6αβ=ca=61=6
  3. Let SS and PP respectively be the sum and products of zeros of the required polynomial.
  4. S=α2β2+β2α2=α4+β4α2β2=(α2+β2)22α2β2(αβ)2=((α+β)22αβ)22(αβ)2(αβ)2S=α2β2+β2α2=α4+β4α2β2=(α2+β2)22α2β2(αβ)2=((α+β)22αβ)22(αβ)2(αβ)2
  5. S=((1)22(6))22(6)2(6)2=9736S=((1)22(6))22(6)2(6)2=9736
  6. P=(α2β2)(β2α2)=1P=(α2β2)(β2α2)=1
  7. Thus, required polynomial will be k(x2Sx+P)=k(x29736x+1)[ Where k is a constant ]

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