If alpha and beta are the two 0's of the polynomial 21y 2 -y-2. find a quadratic polynomial whose 0's are 2alpha& 2beta
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α,β are the zeros of 21y²-y-2=0
then, α+β=-(-1/21)=1/21 and α×β=-2/21
i) now, 2α+2β=2(α+β)=2(1/21)=2/21 and
2α×2β=4αβ=4×(-2/21)=-8/21
the equation having zeros 2α and 2β is:
x²-(sum of the zeros)x+product of the zeros=0
or, x²-(2/21)x+(-8/21)=0
or, x²-2x/21-8/21=0
or, 21x²-2x-8=0
then, α+β=-(-1/21)=1/21 and α×β=-2/21
i) now, 2α+2β=2(α+β)=2(1/21)=2/21 and
2α×2β=4αβ=4×(-2/21)=-8/21
the equation having zeros 2α and 2β is:
x²-(sum of the zeros)x+product of the zeros=0
or, x²-(2/21)x+(-8/21)=0
or, x²-2x/21-8/21=0
or, 21x²-2x-8=0
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Here is your answer user.
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