Geography, asked by ssrndl22, 9 months ago

If a.b are roots of ax2

+ bx + c = 0 then a^3+b^3 ​

Answers

Answered by samruddipatil9
1

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.

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