if alpha and beta are the zeros of 2 x square - 5 x + 3 find the value of Alpha square plus beta square second one upon alpha + one upon beta third Alpha Cube + beta cube IV Alpha upon beta + beta upon Alpha.
Answers
If alpha and beta are the zeroes of a quadratic polynomial, 3x^2 +5x +7, then what is /alpha cube +1 /beta cube?
α+β=-5/3
αβ=7/3
Also
(α+β)^2=α^2+β^2+2αβ
Put values to get α^2+β^2
Now
1/α^3 + 1/β^3 =(α^3+β^3)/α^3β^3
={(α+β)(α^2+β^2-αβ)}/(αβ)^3
Now just put the values to get it
1.2k views
Let a,b be the roots
Then a+b=-5/3
ab=7/3
(1/a^3)+(1/b^3)=a^3+b^3/(ab)^3. Eqn (1)
(a+b)^3=a^3+b^3+3ab(a+b)
Hence a^3+b^3=(-5/3)^3–3×7/3×(-5/3)
440/27
Hence from eqn 1
(1/a^3)+(1/b^3)=440/343
33 views
Let 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 , Answer
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