If a.b are roots of ax2
+ bx + c = 0 then a^3+b^3
Answers
Explanation: ax2+bx+c=0 and the roots of the equation is α and β , make new quadratic equation with x1=α3 & x2=β3 .
by the definition quadratic equation ,
(x−α)(x−β)=0
x2−(α+β)x+αβ=0
same form by :
x2+bax+ca=0 -> Both equation devided by a.
and we can conclude that :
(α+β)=−ba
αβ=ca
so,
in new quadratic equation
(x−α3)(x−β3)=0
x2−(α3+β3)x+(αβ)3=0
.: Let’s see
(α+β)3=α3+3α2β+3αβ2+\bet … (more)
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Herberth Diestro, I'm learning algorithms
Answered July 18, 2016
ax^2 + bx + c = 0,
x^2 + (b/a) x + (c/a) = 0
If α, β are roots of the equation ax^2 + bx + c = 0, then:
(x - α) (x - β) = x^2 + (b/a) x + (c/a)
from this we have:
α + β = - b/a . . . . . . . [1]
αβ = c/a . . . . . . . . . . . [2]
Now,
(x - α^3) (x - β^3) =
x^2 - x (α^3 + β^3) + (αβ)^3 =
x^2 - x [ (α+β)^3 - 3(α^2)β - 3α(β^2) ] + (αβ)^3 =
x^2 - x [ (α+β)^3 - 3αβ(α+β) ] + (αβ)^3 =
Replacing [1] and [2] in previous equation, we have:
x^2 - x [ (-b/a)^3 - 3(c/a)(-b/a) ] + (c/a)^3 =
Multiply by (a^3) you will have:
(a^3) x^2 + (b^3 - 3abc) x + (c^3) = 0 ……….. [the answer!!!]
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Harsh Dwivedi, former Student
Answered December 9, 2019
Let f(x) be a polynomial of degree 2 and its root be,α,β
Then it can be expressed in form of
x2−(α+β)x+αβ
Same thing can be application to equation
⟹α+β=−ba
⟹αβ=ca
⟹(α)3+(β)3=(α+β)((α)2−αβ+(β)2)
⟹(α)3+(β)3=(α+β)((α+β)2–3αβ)
⟹(α)3+(β)3=(−ba)(b2a2−3ca)
[math]\implies (\alpha)^3+(\beta)^3=\dfrac{-b}{a}\left(\dfrac{b^2[/math] … (more)
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Related Questions
More Answers Below
Mathematics: If α, β are the roots of ax^2 + bx + c = 0, then -1/α , -1/β are the roots of what equation?
If alpha and beta are roots of the quadratic equation ax²+bx+c=0 then form an equation whose roots are alpha+1/beta, beta+1/alpha?
If α, β are the roots of the equation ax2+bx+c=0, then what is the quadratic equation whose roots are 2α+3β and 3α+2β?
Ved Prakash Sharma, former Lecturer at Sbm Inter College, Rishikesh (1971-2007)
Answered December 30, 2017
(Let alpha=p ,and beta=q.)
p+q=-b/a……………(1)
p.q=c/a……………….(2)
Sum of roots=p^3+q^3=(p+q)^3–3pq(p+q)
=(-b/a)^3–3(c/a).(-b/a)=-b^3/a^3+3bc/a^2
=(3abc-b^3)/a^3
Product of roots=p^3.q^3=(p.q)^3=(c/a)^3=c^3/a^3
Required equation is:-
x^2-(sum of roots).x+product of roots=0
x^2-(3abc-c^3).x/a^3 +c^3/a^3=0
or a^3.x^2-(3abc-c^3).x+c^3=0 , Answer.