Solving $a^3+a^2|a+x|+|a^2x+1|=1$

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Find all values of $a$ for which the equation $$a^3+a^2|a+x|+|a^2x+1|=1$$ has no less than four different integer solution.

My attempts:

If we rearrange it as $$|a^2x+1|+|a^3+a^2x|=(a^2x+1)-(a^3+a^2x)$$

Hence it requires us to solve following system of equation,
begincasesa^2x+1&geq0\ a^3+a^2x&leq0endcases

And here I'm paused. Please help

algebra-precalculus inequality systems-of-equations integers

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asked Jul 20 at 4:05

mnulb

1,326620

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
  – Cornman
  Jul 20 at 4:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman the interval of $a$ such that $x$ can take four integer solutions.
  – mnulb
  Jul 20 at 4:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
  – Christopher Marley
  Jul 20 at 5:09
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Find all values of $a$ for which the equation $$a^3+a^2|a+x|+|a^2x+1|=1$$ has no less than four different integer solution.

My attempts:

If we rearrange it as $$|a^2x+1|+|a^3+a^2x|=(a^2x+1)-(a^3+a^2x)$$

Hence it requires us to solve following system of equation,
begincasesa^2x+1&geq0\ a^3+a^2x&leq0endcases

And here I'm paused. Please help

algebra-precalculus inequality systems-of-equations integers

share|cite|improve this question

asked Jul 20 at 4:05

mnulb

1,326620

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
  – Cornman
  Jul 20 at 4:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman the interval of $a$ such that $x$ can take four integer solutions.
  – mnulb
  Jul 20 at 4:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
  – Christopher Marley
  Jul 20 at 5:09
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Find all values of $a$ for which the equation $$a^3+a^2|a+x|+|a^2x+1|=1$$ has no less than four different integer solution.

My attempts:

If we rearrange it as $$|a^2x+1|+|a^3+a^2x|=(a^2x+1)-(a^3+a^2x)$$

Hence it requires us to solve following system of equation,
begincasesa^2x+1&geq0\ a^3+a^2x&leq0endcases

And here I'm paused. Please help

algebra-precalculus inequality systems-of-equations integers

share|cite|improve this question

asked Jul 20 at 4:05

mnulb

1,326620

Find all values of $a$ for which the equation $$a^3+a^2|a+x|+|a^2x+1|=1$$ has no less than four different integer solution.

My attempts:

If we rearrange it as $$|a^2x+1|+|a^3+a^2x|=(a^2x+1)-(a^3+a^2x)$$

Hence it requires us to solve following system of equation,
begincasesa^2x+1&geq0\ a^3+a^2x&leq0endcases

And here I'm paused. Please help

algebra-precalculus inequality systems-of-equations integers

share|cite|improve this question

asked Jul 20 at 4:05

mnulb

1,326620

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 4:05

mnulb

1,326620

asked Jul 20 at 4:05

mnulb

1,326620

asked Jul 20 at 4:05

mnulb

1,326620

1,326620

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
  – Cornman
  Jul 20 at 4:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman the interval of $a$ such that $x$ can take four integer solutions.
  – mnulb
  Jul 20 at 4:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
  – Christopher Marley
  Jul 20 at 5:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
  – Cornman
  Jul 20 at 4:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman the interval of $a$ such that $x$ can take four integer solutions.
  – mnulb
  Jul 20 at 4:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
  – Christopher Marley
  Jul 20 at 5:09
  

  
  
  
  

Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
– Cornman
Jul 20 at 4:47

Are you searching for pairs $(a,x)$ which solve the equation? Or are you asked to solve it for $x$?
– Cornman
Jul 20 at 4:47

@Cornman the interval of $a$ such that $x$ can take four integer solutions.
– mnulb
Jul 20 at 4:56

@Cornman the interval of $a$ such that $x$ can take four integer solutions.
– mnulb
Jul 20 at 4:56

desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
– Christopher Marley
Jul 20 at 5:09

desmos.com/calculator/itnds5xkbw I wish I could help more, but play with the 'a' slider.
– Christopher Marley
Jul 20 at 5:09

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

Consider the case $a=0$ separately, because division by $a^2$ will feature in the manipulations that follow. If $a$ is as such, the inequalities reduce to the trivial $1ge0$ and $0le0$, so certainly the original equation has at least four integer solutions here.

Now assume $ane0$; the first equation can be manipulated as
$$a^2xge-1implies xge-frac1a^2$$
and the second as
$$a^2(a+x)le0implies a+xle0implies xle-a$$
The two bounds meet when $a=1$ and the lower bound exceeds the upper bound when $a>1$, so solutions for $x$ exist only if $ale1$. To get the intervals where $x$ has at least four integer solutions, we consider three intervals arising from the partition of $(-infty,1]$:

• $ain(0,1]$. In this range, $-1$ is a solution for $x$, with lower negative integers appearing as solutions as $a$ approaches 0. We seek the point where $-4$ becomes a solution, which is where $-frac1a^2=-4$ or $a=frac12$.


• $ain[-1,0)$. At $a=-1$, there are only three integer solutions for $x$. Above this, the highest integer solution is 0, so we seek the point where $-3$ becomes a solution as $-frac1a^2$ decreases. This point is $a=-frac1sqrt3$.


• $ain(-infty,-1]$. Downwards of $-1$, the lowest integer solution of $x$ is 0. We seek the point where 3 becomes a solution as $-x$ increases: $a=-3$.

Combining these pieces, we see that the original equation has at least four integer solutions when $ain(-infty,-3]cup[-1/sqrt3,1/2]$.

share|cite|improve this answer

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Consider the case $a=0$ separately, because division by $a^2$ will feature in the manipulations that follow. If $a$ is as such, the inequalities reduce to the trivial $1ge0$ and $0le0$, so certainly the original equation has at least four integer solutions here.

Now assume $ane0$; the first equation can be manipulated as
$$a^2xge-1implies xge-frac1a^2$$
and the second as
$$a^2(a+x)le0implies a+xle0implies xle-a$$
The two bounds meet when $a=1$ and the lower bound exceeds the upper bound when $a>1$, so solutions for $x$ exist only if $ale1$. To get the intervals where $x$ has at least four integer solutions, we consider three intervals arising from the partition of $(-infty,1]$:

• $ain(0,1]$. In this range, $-1$ is a solution for $x$, with lower negative integers appearing as solutions as $a$ approaches 0. We seek the point where $-4$ becomes a solution, which is where $-frac1a^2=-4$ or $a=frac12$.


• $ain[-1,0)$. At $a=-1$, there are only three integer solutions for $x$. Above this, the highest integer solution is 0, so we seek the point where $-3$ becomes a solution as $-frac1a^2$ decreases. This point is $a=-frac1sqrt3$.


• $ain(-infty,-1]$. Downwards of $-1$, the lowest integer solution of $x$ is 0. We seek the point where 3 becomes a solution as $-x$ increases: $a=-3$.

Combining these pieces, we see that the original equation has at least four integer solutions when $ain(-infty,-3]cup[-1/sqrt3,1/2]$.

share|cite|improve this answer

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

add a comment | 

up vote
3
down vote



accepted

Consider the case $a=0$ separately, because division by $a^2$ will feature in the manipulations that follow. If $a$ is as such, the inequalities reduce to the trivial $1ge0$ and $0le0$, so certainly the original equation has at least four integer solutions here.

Now assume $ane0$; the first equation can be manipulated as
$$a^2xge-1implies xge-frac1a^2$$
and the second as
$$a^2(a+x)le0implies a+xle0implies xle-a$$
The two bounds meet when $a=1$ and the lower bound exceeds the upper bound when $a>1$, so solutions for $x$ exist only if $ale1$. To get the intervals where $x$ has at least four integer solutions, we consider three intervals arising from the partition of $(-infty,1]$:

• $ain(0,1]$. In this range, $-1$ is a solution for $x$, with lower negative integers appearing as solutions as $a$ approaches 0. We seek the point where $-4$ becomes a solution, which is where $-frac1a^2=-4$ or $a=frac12$.


• $ain[-1,0)$. At $a=-1$, there are only three integer solutions for $x$. Above this, the highest integer solution is 0, so we seek the point where $-3$ becomes a solution as $-frac1a^2$ decreases. This point is $a=-frac1sqrt3$.


• $ain(-infty,-1]$. Downwards of $-1$, the lowest integer solution of $x$ is 0. We seek the point where 3 becomes a solution as $-x$ increases: $a=-3$.

Combining these pieces, we see that the original equation has at least four integer solutions when $ain(-infty,-3]cup[-1/sqrt3,1/2]$.

share|cite|improve this answer

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Consider the case $a=0$ separately, because division by $a^2$ will feature in the manipulations that follow. If $a$ is as such, the inequalities reduce to the trivial $1ge0$ and $0le0$, so certainly the original equation has at least four integer solutions here.

Now assume $ane0$; the first equation can be manipulated as
$$a^2xge-1implies xge-frac1a^2$$
and the second as
$$a^2(a+x)le0implies a+xle0implies xle-a$$
The two bounds meet when $a=1$ and the lower bound exceeds the upper bound when $a>1$, so solutions for $x$ exist only if $ale1$. To get the intervals where $x$ has at least four integer solutions, we consider three intervals arising from the partition of $(-infty,1]$:

• $ain(0,1]$. In this range, $-1$ is a solution for $x$, with lower negative integers appearing as solutions as $a$ approaches 0. We seek the point where $-4$ becomes a solution, which is where $-frac1a^2=-4$ or $a=frac12$.


• $ain[-1,0)$. At $a=-1$, there are only three integer solutions for $x$. Above this, the highest integer solution is 0, so we seek the point where $-3$ becomes a solution as $-frac1a^2$ decreases. This point is $a=-frac1sqrt3$.


• $ain(-infty,-1]$. Downwards of $-1$, the lowest integer solution of $x$ is 0. We seek the point where 3 becomes a solution as $-x$ increases: $a=-3$.

Combining these pieces, we see that the original equation has at least four integer solutions when $ain(-infty,-3]cup[-1/sqrt3,1/2]$.

share|cite|improve this answer

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

Consider the case $a=0$ separately, because division by $a^2$ will feature in the manipulations that follow. If $a$ is as such, the inequalities reduce to the trivial $1ge0$ and $0le0$, so certainly the original equation has at least four integer solutions here.

Now assume $ane0$; the first equation can be manipulated as
$$a^2xge-1implies xge-frac1a^2$$
and the second as
$$a^2(a+x)le0implies a+xle0implies xle-a$$
The two bounds meet when $a=1$ and the lower bound exceeds the upper bound when $a>1$, so solutions for $x$ exist only if $ale1$. To get the intervals where $x$ has at least four integer solutions, we consider three intervals arising from the partition of $(-infty,1]$:

• $ain(0,1]$. In this range, $-1$ is a solution for $x$, with lower negative integers appearing as solutions as $a$ approaches 0. We seek the point where $-4$ becomes a solution, which is where $-frac1a^2=-4$ or $a=frac12$.


• $ain[-1,0)$. At $a=-1$, there are only three integer solutions for $x$. Above this, the highest integer solution is 0, so we seek the point where $-3$ becomes a solution as $-frac1a^2$ decreases. This point is $a=-frac1sqrt3$.


• $ain(-infty,-1]$. Downwards of $-1$, the lowest integer solution of $x$ is 0. We seek the point where 3 becomes a solution as $-x$ increases: $a=-3$.

Combining these pieces, we see that the original equation has at least four integer solutions when $ain(-infty,-3]cup[-1/sqrt3,1/2]$.

share|cite|improve this answer

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 5:35

edited Jul 20 at 5:35

edited Jul 20 at 5:35

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

answered Jul 20 at 5:15

Parcly Taxel

33.6k136588

33.6k136588

add a comment | 

add a comment | 

 

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