if alpha and beta are the zeros of the polynomial X square + 7x +7 then find the value of :-
(part 1) Alpha square plus beta square
( part 2 )one upon alpha plus one upon beta
Answers
☺✌◦•●◉✿ α+β=-5/3
αβ=7/3
ᴀˡˢᵒ
(α+β)^2=α^2+β^2+2αβ
ᴘᵘᵗ ᵛᵃˡᵘᵉˢ ᵗᵒ ᵍᵉᵗ α^2+β^2
ɴᵒʷ
1/α^3 + 1/β^3 =(α^3+β^3)/α^3β^3
={(α+β)(α^2+β^2-αβ)}/(αβ)^3
ɴᵒʷ ʲᵘˢᵗ ᵖᵘᵗ ᵗʰᵉ ᵛᵃˡᵘᵉˢ ᵗᵒ ᵍᵉᵗ ⁱᵗ
1.2ᵏ ᵛⁱᵉʷˢ
ʟᵉᵗ ᵃ,ᵇ ᵇᵉ ᵗʰᵉ ʳᵒᵒᵗˢ
ᴛʰᵉⁿ ᵃ+ᵇ=-5/3
ᵃᵇ=7/3
(1/ᵃ^3)+(1/ᵇ^3)=ᵃ^3+ᵇ^3/(ᵃᵇ)^3. ᴇᵠⁿ (1)
(ᵃ+ᵇ)^3=ᵃ^3+ᵇ^3+3ᵃᵇ(ᵃ+ᵇ)
ʜᵉⁿᶜᵉ ᵃ^3+ᵇ^3=(-5/3)^3–3×7/3×(-5/3)
440/27
ʜᵉⁿᶜᵉ ᶠʳᵒᵐ ᵉᵠⁿ 1
(1/ᵃ^3)+(1/ᵇ^3)=440/343
33 ᵛⁱᵉʷˢ
ʟᵉᵗ ᵃˡᵖʰᵃ=ᵃ , ᵇᵉᵗᵃ=ᵇ.
ᵃ ᵃⁿᵈ ᵇ ᵃʳᵉ ᵗʰᵉ ᶻᵉʳᵒˢ ᵒᶠ 3ˣ^2+5ˣ+7 ,
ᵃ+ᵇ=-5/3………(1)
ᵃ.ᵇ=7/3…………..(2)
1/ᵃ^3+1/ᵇ^3=(1/ᵃ^3.ᵇ^3)[ᵇ^3+ᵃ^3]
=(1/ᵃ^3.ᵇ^3)(ᵃ+ᵇ)(ᵃ^2-ᵃᵇ+ᵇ^2)
=(1/ᵃ^3.ᵇ^3)(ᵃ+ᵇ)[(ᵃ+ᵇ)^2–3ᵃ.ᵇ]
=(3/7)^3.(-5/3)[25/9–3.7/3]
=(27/343)(-5/3)(-38/9)
=(27×5×38)/(343×3×9)
=190/343 , ᴀⁿˢʷᵉʳ
✿◉●•◦
X^2 + 7x +7
a=1,b=7 and c=7
(1) alpha ^2 + betha ^2.
(A) since (alpha + betha )=-b/a
(alpha + betha) = -7/1
(alpha + betha)=-7
(alpha + betha)^2=(-b/a)^2
sine alpha +betha = -7
so ,( -7)^2 -2( alpha ×betha)=alpha ^2 + betha ^2.
so 49 - 2(c/a)=. alpha ^2 + betha ^2.
49 - 2(7)=. alpha ^2 + betha ^2.
49-14 =. alpha ^2 + betha ^2
35. =alpha ^2 + betha ^2.
there fore
alpha ^2 + betha ^2. =35
(2) 1/alpha + 1/betha (cross multiplication)
alpha + betha /alpha×betha
so , -b/a. ÷c/a
-7/7
-1
there fore
1/alpha + 1/betha =-1
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