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How do I find the solutions of the following equation?

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2

1

$begingroup$

How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$

It has 5 solutions, 4 positive and 1 negative. The graphs are these.

How do I compute the values of these roots manually?

algebra-precalculus

share|cite|improve this question

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3157637/…
  $endgroup$
  – lab bhattacharjee
  2 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$

It has 5 solutions, 4 positive and 1 negative. The graphs are these.

How do I compute the values of these roots manually?

algebra-precalculus

share|cite|improve this question

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3157637/…
  $endgroup$
  – lab bhattacharjee
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$

It has 5 solutions, 4 positive and 1 negative. The graphs are these.

How do I compute the values of these roots manually?

algebra-precalculus

share|cite|improve this question

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 3 hours ago

Maria Mazur

48.6k1260121

edited 3 hours ago

Maria Mazur

48.6k1260121

asked 3 hours ago

Namami ShankerNamami Shanker

111

New contributor

Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

Namami ShankerNamami Shanker

111

4 Answers
4

active

oldest

votes

5

$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.

Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$

So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.

Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.

share|cite|improve this answer

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

$endgroup$

add a comment | 

2

$begingroup$

So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

$endgroup$

add a comment | 

0

$begingroup$

We get easy that $$x=2$$ is one solution.
Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
Can you finish?
Hint: $$x=3$$ and $$x=1$$ are also solutions.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you sir.
  $endgroup$
  – Namami Shanker
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not give all of the solutions.
  $endgroup$
  – Peter Foreman
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
  $endgroup$
  – Robert Israel
  3 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint

Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)

share|cite|improve this answer

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

$endgroup$

add a comment | 

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5

$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.

Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$

So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.

Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.

share|cite|improve this answer

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

$endgroup$

add a comment | 

5

$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.

Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$

So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.

Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.

share|cite|improve this answer

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

$endgroup$

add a comment | 

$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.

Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$

So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.

Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.

share|cite|improve this answer

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

$endgroup$

share|cite|improve this answer

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

edited 1 hour ago

Moo

5,64131020

edited 1 hour ago

Moo

5,64131020

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

answered 3 hours ago

Maria MazurMaria Mazur

48.6k1260121

2

$begingroup$

So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

$endgroup$

add a comment | 

2

$begingroup$

So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

$endgroup$

add a comment | 

$begingroup$

So rearranging gives
$$|x-2|^10x^2-1-|x-2|^3x=0$$
$$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

answered 3 hours ago

Peter ForemanPeter Foreman

4,2721216

0

$begingroup$

We get easy that $$x=2$$ is one solution.
Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
Can you finish?
Hint: $$x=3$$ and $$x=1$$ are also solutions.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you sir.
  $endgroup$
  – Namami Shanker
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not give all of the solutions.
  $endgroup$
  – Peter Foreman
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
  $endgroup$
  – Robert Israel
  3 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

We get easy that $$x=2$$ is one solution.
Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
Can you finish?
Hint: $$x=3$$ and $$x=1$$ are also solutions.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thank you sir.
  $endgroup$
  – Namami Shanker
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not give all of the solutions.
  $endgroup$
  – Peter Foreman
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
  $endgroup$
  – Robert Israel
  3 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

We get easy that $$x=2$$ is one solution.
Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
Can you finish?
Hint: $$x=3$$ and $$x=1$$ are also solutions.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

answered 3 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

0

$begingroup$

Hint

Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)

share|cite|improve this answer

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

$endgroup$

add a comment | 

0

$begingroup$

Hint

Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)

share|cite|improve this answer

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

$endgroup$

add a comment | 

$begingroup$

Hint

Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)

share|cite|improve this answer

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040

answered 2 hours ago

Mostafa AyazMostafa Ayaz

18k31040