Question

If a,b are the zeroes of the polynomial 21y^2-y-2, find a quadratic polynomial whose zeroes are 2a and 2b

Answer 1

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

Answer 2

Step-by-step explanation:

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