Please explain in a way it can be understood easily.. My math is kinda bad...
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α,β are the zeros of x²-2x+3=0
then, α+β=-(-2/1)=2 and α×β=3/1=3
i) now, (α+2)+(β+2)=(α+β)+4=2+4=6 and
(α+2)(β+2)=αβ+2β+2α+4=3+2(α+β)+4=7+2.2=7+4=11
the equation having zeros (α+2) and (β+2) is:
x²-(sum of the zeros)x+product of the zeros=0
or, x²-6x+11=0
Hope you can do the rest of two problems.
then, α+β=-(-2/1)=2 and α×β=3/1=3
i) now, (α+2)+(β+2)=(α+β)+4=2+4=6 and
(α+2)(β+2)=αβ+2β+2α+4=3+2(α+β)+4=7+2.2=7+4=11
the equation having zeros (α+2) and (β+2) is:
x²-(sum of the zeros)x+product of the zeros=0
or, x²-6x+11=0
Hope you can do the rest of two problems.
iamshru:
hey thanks a lot.. ☺☺
welcome :)
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