Mixing
Finding a quadratic equation using roots
Clash Royale CLAN TAG#URR8PPP
up vote
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If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_1/2$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_1/2=large-bpmsqrt4acover2a$
and from here:
$$beginalignx_1^3&=bigg(-b+sqrt4acover2abigg)^3\&=(sqrt4ac-b)^2(sqrt4ac-b)over8a^3\&=(4ac-2bsqrt4ac+b^2)(sqrt4ac-b)over8a^3\&=4acsqrt4ac-4abc-8abc-2b^2sqrt4ac+b^2sqrt4ac-b^3over8a^3\&=4acsqrt4ac-12abc-b^2sqrt4ac-b^3over8a^3endalign$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
quadratics
asked 48 mins ago
Aleksa
25912
add a comment |Â
up vote
2
down vote
favorite
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_1/2$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_1/2=large-bpmsqrt4acover2a$
and from here:
$$beginalignx_1^3&=bigg(-b+sqrt4acover2abigg)^3\&=(sqrt4ac-b)^2(sqrt4ac-b)over8a^3\&=(4ac-2bsqrt4ac+b^2)(sqrt4ac-b)over8a^3\&=4acsqrt4ac-4abc-8abc-2b^2sqrt4ac+b^2sqrt4ac-b^3over8a^3\&=4acsqrt4ac-12abc-b^2sqrt4ac-b^3over8a^3endalign$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
quadratics
asked 48 mins ago
Aleksa
25912
Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_1/2$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_1/2=large-bpmsqrt4acover2a$
and from here:
$$beginalignx_1^3&=bigg(-b+sqrt4acover2abigg)^3\&=(sqrt4ac-b)^2(sqrt4ac-b)over8a^3\&=(4ac-2bsqrt4ac+b^2)(sqrt4ac-b)over8a^3\&=4acsqrt4ac-4abc-8abc-2b^2sqrt4ac+b^2sqrt4ac-b^3over8a^3\&=4acsqrt4ac-12abc-b^2sqrt4ac-b^3over8a^3endalign$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
quadratics
asked 48 mins ago
Aleksa
25912
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_1/2$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_1/2=large-bpmsqrt4acover2a$
and from here:
$$beginalignx_1^3&=bigg(-b+sqrt4acover2abigg)^3\&=(sqrt4ac-b)^2(sqrt4ac-b)over8a^3\&=(4ac-2bsqrt4ac+b^2)(sqrt4ac-b)over8a^3\&=4acsqrt4ac-4abc-8abc-2b^2sqrt4ac+b^2sqrt4ac-b^3over8a^3\&=4acsqrt4ac-12abc-b^2sqrt4ac-b^3over8a^3endalign$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
quadratics
quadratics
asked 48 mins ago
Aleksa
25912
asked 48 mins ago
Aleksa
25912
asked 48 mins ago
Aleksa
25912
asked 48 mins ago
Aleksa
25912
asked 48 mins ago
Aleksa
25912
25912
Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago
add a comment |Â
Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago
Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago
Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
answered 46 mins ago
gimusi
82.6k74091
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
add a comment |Â
up vote
3
down vote
Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
answered 43 mins ago
ajotatxe
51.5k23286
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
answered 46 mins ago
gimusi
82.6k74091
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
add a comment |Â
up vote
4
down vote
accepted
HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
answered 46 mins ago
gimusi
82.6k74091
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
answered 46 mins ago
gimusi
82.6k74091
HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
answered 46 mins ago
gimusi
82.6k74091
answered 46 mins ago
gimusi
82.6k74091
answered 46 mins ago
gimusi
82.6k74091
answered 46 mins ago
gimusi
82.6k74091
82.6k74091
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
add a comment |Â
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
Is this correct considering there's $a$? And I made a mistake in my question saying $(x-x_1^3)(x-x_2^3)$, it should be $(ax-x_1^3)(ax-x_2^3)$ I think
– Aleksa
38 mins ago
1
1
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
@Aleksa As noticed we can divide wlog by $a$ and then consider an equation in the form x^2+Bx+C, or consider $$a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2$$ $$a(x-x_1^3)(x-x_2^3)=ax^2-a(x_1^3+x_2^3)x+ax_1^3x_2^3$$ and proceed comapring the terms.
– gimusi
34 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
Yeah, you're right, I got $a^3x^2+b(b^2-3ac)x+c^3=0$, should be correct I think
– Aleksa
31 mins ago
1
1
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
@Aleksa Yes I obtain the same result!
– gimusi
26 mins ago
add a comment |Â
up vote
3
down vote
Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
answered 43 mins ago
ajotatxe
51.5k23286
add a comment |Â
up vote
3
down vote
Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
answered 43 mins ago
ajotatxe
51.5k23286
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
answered 43 mins ago
ajotatxe
51.5k23286
Let $B=b/a$ and $C=c/a$. Then $x_1$ and $x_2$ are the roots of $x^2+Bx+C$. Moreover, $x_1+x_2=-B$ and $x_1x_2=C$.
The roots of the polynomial
$$x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
are $x_1^3$ and $x_2^3$. But $x_1^3x_2^3=C^3$ and $x_1^3+x_2^3=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)=-B(B^2-3C)$
answered 43 mins ago
ajotatxe
51.5k23286
answered 43 mins ago
ajotatxe
51.5k23286
answered 43 mins ago
ajotatxe
51.5k23286
answered 43 mins ago
ajotatxe
51.5k23286
51.5k23286
add a comment |Â
add a comment |Â
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Hint: Just replace $x$ with $sqrt[3] x$ in the given quadratic to obtain required quadratic.
– Manthanein
45 mins ago