If α and β are the zeros of polynomial x2−x−6, find a polynomial whose zeros are α2β2 and β2α2.
Answer:
k(x2−9736x+1)
- On comparing the polynomial x2−x−6, with the standard form ax2+bx+c=0, we get:
a=1b=−1c=−6 - Since, sum of zeros α+β=−ba=11=1
Product of zeros αβ=ca=−61=−6 - Let S and P respectively be the sum and products of zeros of the required polynomial.
- S=α2β2+β2α2=α4+β4α2β2=(α2+β2)2−2α2β2(αβ)2=((α+β)2−2αβ)2−2(αβ)2(αβ)2
- S=((1)2−2(−6))2−2(−6)2(−6)2=9736
- P=(α2β2)(β2α2)=1
- Thus, required polynomial will be k(x2−Sx+P)=k(x2−9736x+1)[ Where k is a constant ]