If $α, β$ $alpha,beta$ are the roots of equation $x^{2} - p x + q = 0$ $x^2-px+q=0$ then find the quadratic equation the roots of which are $(α^{2} - β^{2}) (α^{3} - β^{3}) & α^{3} β^{2} + α^{2} β^{3}$ $(alpha^2-beta^2)(alpha^3-beta^3) & alpha^3 beta^2+ alpha^2 beta^3$ ?

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If $α, β$ $alpha,beta$ are the roots of equation $x^{2} - p x + q = 0$ $x^2-px+q=0$ then find the quadratic equation the roots of which are $(α^{2} - β^{2}) (α^{3} - β^{3}) & α^{3} β^{2} + α^{2} β^{3}$ $(alpha^2-beta^2)(alpha^3-beta^3) & alpha^3 beta^2+ alpha^2 beta^3$ ?

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Answer 1 · 2017-07-20T09:36:45

$x^{2} - (p^{5} - 5 p^{3} q + 5 p q^{2}) x + (p^{6} q^{2} - 5 p^{4} q^{3} + 4 p^{2} q^{4}) = 0$ $x^2-(p^5-5p^3q+5pq^2)x + (p^6q^2-5p^4q^3+4p^2q^4) = 0$

Explanation:

Given:

$x^{2} - p x + q = (x - α) (x - β) = x^{2} - (α + β) x + α β$ $x^2-px+q = (x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta$

We have:

${ (alpha+beta = p), (alphabeta = q) :}$

Hence:

$(alpha^2-beta^2)(alpha^3-beta^3) = (alpha-beta)^2(alpha+beta)(alpha^2+alphabeta+beta^2)$

$color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = (alpha^2-2alphabeta+beta^2)(alpha+beta)(alpha^2+alphabeta+beta^2)$

$color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = ((alpha+beta)^2-4alphabeta)(alpha+beta)((alpha+beta)^2-alphabeta)$

$color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = (p^2-4q)p(p^2-q)$

$color(white)((alpha^2-beta^2)(alpha^3-beta^3)) = p^5-5p^3q+4pq^2$

and:

$alpha^3beta^2+alpha^2beta^3 = (alpha+beta)(alphabeta)^2 = pq^2$

So the monic quadratic equation with the these two roots can be written:

$0 = x^2-((p^5-5p^3q+4pq^2) + pq^2)x + (p^5-5p^3q+4pq^2)pq^2$

$color(white)(0) = x^2-(p^5-5p^3q+5pq^2)x + (p^6q^2-5p^4q^3+4p^2q^4)$

Answer 2 · 2017-07-20T09:59:38

Equation is
$x^2-(p^5-5p^3q+5pq^2)x+(p^6q^2-5p^4q^3+4p^2q^4)=0$

Explanation:

As $alpha$ and $beta$ are roots of $x^2-px+q=0$

$alpha+beta=p$ and $alphabeta=q$

Therefore $alpha^2+beta^2=p^2-2q$

$alpha^3+beta^3=(alpha+beta)(alpha^2-alphabeta+beta^2)$

= $p(p^2-2q-q)=p(p^2-3q)=p^3-3pq$

and $(alpha^2+beta^2)(alpha^3+beta^3)=alpha^5+beta^5+alpha^2beta^2(alpha+beta)$

or $(p^2-2q)(p^3-3pq)=alpha^5+beta^5+pq^2$

or $alpha^5+beta^5=(p^2-2q)(p^3-3pq)-pq^2$

The quadratic equation whose roots are given is

is $x^2-("sum of roots")x+("product of roots")=0$

Sum of roots is $(alpha^2-beta^2)(alpha^3-beta^3)+alpha^3beta^2+alpha^2beta^3$

= $alpha^5+beta^5-alpha^2beta^3-beta^2alpha^3+alpha^3beta^2+alpha^2beta^3=alpha^5+beta^5$

= $(p^2-2q)(p^3-3pq)-pq^2=p^5-5p^3q+5pq^2$

and product of roots is $(alpha^5+beta^5-alpha^2beta^3-beta^2alpha^3)(alpha^3beta^2+alpha^2beta^3)$

= $(p^5-5p^3q+5pq^2-pq^2)(pq^2)$

= $p^6q^2-5p^4q^3+4p^2q^4$

Hence equation is

$x^2-(p^5-5p^3q+5pq^2)x+(p^6q^2-5p^4q^3+4p^2q^4)=0$

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If $α, β$ $alpha,beta$ are the roots of equation $x^{2} - p x + q = 0$ $x^2-px+q=0$ then find the quadratic equation the roots of which are $(α^{2} - β^{2}) (α^{3} - β^{3}) & α^{3} β^{2} + α^{2} β^{3}$ $(alpha^2-beta^2)(alpha^3-beta^3) & alpha^3 beta^2+ alpha^2 beta^3$ ?

2 Answers

Explanation:

Explanation:

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