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Why isn't the definition of absolute value applied when squaring a radical containing a variable?

Different ways of defining Absolute ValueAdding $2$ absolute values together: $|x+2| + |x-3| =5.$Why does $|x_1| = |x_2| implies x_1 = pm x_2$Why does the integral $intfrac1x+idx$ not require the absolute value in the logarithm?The Definition of the Absolute ValueSpivak's Calculus 4th ed: Absolute value definition p. 11Solving Equations with Roots and Absolute ValuesAbsolute Value. Real AnalysisSolving Radical InequalitiesHow to get rid of the absolute value when solving a system of complex equations?

5

$begingroup$

I recently learned about the following definition of absolute value:

$|a| = sqrta^2$

Then I came across a solution to a problem that had the following step:

$5 geq sqrt5 - x$

In order to proceed, we had to square both sides:

$5^2 geq (sqrt5 - x)^2$

With the aforementioned definition of absolute value in mind, I wrote:

$25 geq |5 - x|$

But the actual solution turned out to be:

$25 geq 5 - x$

I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 27 at 14:30

CalculemusCalculemus

439317

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt $, but after solving the resulting inequality, you need to check for false solutions.
  $endgroup$
  – user647486
  Apr 27 at 14:41
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Actually the solution should be $25 geq 5 - x geq 0 implies -20 le x le 5$
  $endgroup$
  – leonbloy
  2 days ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

I recently learned about the following definition of absolute value:

$|a| = sqrta^2$

Then I came across a solution to a problem that had the following step:

$5 geq sqrt5 - x$

In order to proceed, we had to square both sides:

$5^2 geq (sqrt5 - x)^2$

With the aforementioned definition of absolute value in mind, I wrote:

$25 geq |5 - x|$

But the actual solution turned out to be:

$25 geq 5 - x$

I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 27 at 14:30

CalculemusCalculemus

439317

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt $, but after solving the resulting inequality, you need to check for false solutions.
  $endgroup$
  – user647486
  Apr 27 at 14:41
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Actually the solution should be $25 geq 5 - x geq 0 implies -20 le x le 5$
  $endgroup$
  – leonbloy
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I recently learned about the following definition of absolute value:

$|a| = sqrta^2$

Then I came across a solution to a problem that had the following step:

$5 geq sqrt5 - x$

In order to proceed, we had to square both sides:

$5^2 geq (sqrt5 - x)^2$

With the aforementioned definition of absolute value in mind, I wrote:

$25 geq |5 - x|$

But the actual solution turned out to be:

$25 geq 5 - x$

I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 27 at 14:30

CalculemusCalculemus

439317

$endgroup$

share|cite|improve this question

asked Apr 27 at 14:30

CalculemusCalculemus

439317

share|cite|improve this question

asked Apr 27 at 14:30

CalculemusCalculemus

439317

asked Apr 27 at 14:30

CalculemusCalculemus

439317

asked Apr 27 at 14:30

CalculemusCalculemus

439317

5 Answers
5

active

oldest

votes

6

$begingroup$

From the fact that you can take $sqrt 5-x$ you know that $5-x ge 0$ so you don't need the absolute value signs.

share|cite|improve this answer

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
  $endgroup$
  – steven gregory
  Apr 27 at 14:42
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  @stevengregory - I am curious what this great difference of meaning between "know that" and "must have" is. To me in this context, they are synonymous.
  $endgroup$
  – Paul Sinclair
  Apr 27 at 16:42
  

  
  
  
  

add a comment | 

14

$begingroup$

Indeed, $sqrta^2=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.

share|cite|improve this answer

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed: Writing $ sqrta $ implies that $ a geq 0 $ if $a$ is a real number
  $endgroup$
  – Barranka
  Apr 27 at 20:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know. But see the original context. This was applied to the inequality $5geqslantsqrt5-x$. The OP thought the one could conclude from it that $5^2geqslantlvert5-xrvert$, whereas the correct conclusion is that $5-xgeqslant0$ and that $5^2geqslant5-x$.
  $endgroup$
  – José Carlos Santos
  Apr 27 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But in that case $5-x = |5-x|$, so OP also isn't 'incorrect', per se, just missing an additional detail.
  $endgroup$
  – Hayden
  Apr 27 at 22:03
  

  
  
  
  

add a comment | 

13

$begingroup$

$$left(sqrt aright)^2nesqrta^2.$$

Try with $a=-1$.

share|cite|improve this answer

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

2

$begingroup$

To compare $sqrt5-x$ to $sqrta^2$ you must compare $5-x$ to $a^2$. The problem is that $a^2 ge 0$ while $5-x$ can be any real number. But if you add the restriction $5-x ge 0$, then $25 geq |5 - x|$ becomes

$$text$25 geq 5 - x textand 5-x ge 0$$$

share|cite|improve this answer

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

$endgroup$

add a comment | 

1

$begingroup$

If you look carefully, you'll notice your definition has the square inside
the square root (not outside):
$$|a| = sqrta^2$$
However, in your solution you seem to assume that:
$$sqrta^2 = (sqrta)^2$$
However,
$$sqrta^2 neq (sqrta)^2$$
For example, as others have suggested, if $a = -1$
we have:
$$sqrta^2 = sqrt(-1)^2 = sqrt1 = 1$$
but
$$(sqrta)^2 = (sqrt-1)^2 = i^2 = -1$$

share|cite|improve this answer

answered 2 days ago

user669329user669329

111

New contributor

user669329 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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6

$begingroup$

From the fact that you can take $sqrt 5-x$ you know that $5-x ge 0$ so you don't need the absolute value signs.

share|cite|improve this answer

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
  $endgroup$
  – steven gregory
  Apr 27 at 14:42
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  @stevengregory - I am curious what this great difference of meaning between "know that" and "must have" is. To me in this context, they are synonymous.
  $endgroup$
  – Paul Sinclair
  Apr 27 at 16:42
  

  
  
  
  

add a comment | 

6

$begingroup$

From the fact that you can take $sqrt 5-x$ you know that $5-x ge 0$ so you don't need the absolute value signs.

share|cite|improve this answer

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
  $endgroup$
  – steven gregory
  Apr 27 at 14:42
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  @stevengregory - I am curious what this great difference of meaning between "know that" and "must have" is. To me in this context, they are synonymous.
  $endgroup$
  – Paul Sinclair
  Apr 27 at 16:42
  

  
  
  
  

add a comment | 

$begingroup$

From the fact that you can take $sqrt 5-x$ you know that $5-x ge 0$ so you don't need the absolute value signs.

share|cite|improve this answer

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

$endgroup$

share|cite|improve this answer

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

answered Apr 27 at 14:33

Ross MillikanRoss Millikan

303k24201375

14

$begingroup$

Indeed, $sqrta^2=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.

share|cite|improve this answer

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed: Writing $ sqrta $ implies that $ a geq 0 $ if $a$ is a real number
  $endgroup$
  – Barranka
  Apr 27 at 20:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know. But see the original context. This was applied to the inequality $5geqslantsqrt5-x$. The OP thought the one could conclude from it that $5^2geqslantlvert5-xrvert$, whereas the correct conclusion is that $5-xgeqslant0$ and that $5^2geqslant5-x$.
  $endgroup$
  – José Carlos Santos
  Apr 27 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But in that case $5-x = |5-x|$, so OP also isn't 'incorrect', per se, just missing an additional detail.
  $endgroup$
  – Hayden
  Apr 27 at 22:03
  

  
  
  
  

add a comment | 

14

$begingroup$

Indeed, $sqrta^2=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.

share|cite|improve this answer

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed: Writing $ sqrta $ implies that $ a geq 0 $ if $a$ is a real number
  $endgroup$
  – Barranka
  Apr 27 at 20:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know. But see the original context. This was applied to the inequality $5geqslantsqrt5-x$. The OP thought the one could conclude from it that $5^2geqslantlvert5-xrvert$, whereas the correct conclusion is that $5-xgeqslant0$ and that $5^2geqslant5-x$.
  $endgroup$
  – José Carlos Santos
  Apr 27 at 20:45
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But in that case $5-x = |5-x|$, so OP also isn't 'incorrect', per se, just missing an additional detail.
  $endgroup$
  – Hayden
  Apr 27 at 22:03
  

  
  
  
  

add a comment | 

$begingroup$

Indeed, $sqrta^2=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.

share|cite|improve this answer

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

$endgroup$

share|cite|improve this answer

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

answered Apr 27 at 14:33

José Carlos SantosJosé Carlos Santos

179k24139254

13

$begingroup$

$$left(sqrt aright)^2nesqrta^2.$$

Try with $a=-1$.

share|cite|improve this answer

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

13

$begingroup$

$$left(sqrt aright)^2nesqrta^2.$$

Try with $a=-1$.

share|cite|improve this answer

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

$begingroup$

$$left(sqrt aright)^2nesqrta^2.$$

Try with $a=-1$.

share|cite|improve this answer

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

$endgroup$

share|cite|improve this answer

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

answered Apr 27 at 14:32

Yves DaoustYves Daoust

135k676233

2

$begingroup$

To compare $sqrt5-x$ to $sqrta^2$ you must compare $5-x$ to $a^2$. The problem is that $a^2 ge 0$ while $5-x$ can be any real number. But if you add the restriction $5-x ge 0$, then $25 geq |5 - x|$ becomes

$$text$25 geq 5 - x textand 5-x ge 0$$$

share|cite|improve this answer

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

$endgroup$

add a comment | 

2

$begingroup$

To compare $sqrt5-x$ to $sqrta^2$ you must compare $5-x$ to $a^2$. The problem is that $a^2 ge 0$ while $5-x$ can be any real number. But if you add the restriction $5-x ge 0$, then $25 geq |5 - x|$ becomes

$$text$25 geq 5 - x textand 5-x ge 0$$$

share|cite|improve this answer

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

$endgroup$

add a comment | 

$begingroup$

To compare $sqrt5-x$ to $sqrta^2$ you must compare $5-x$ to $a^2$. The problem is that $a^2 ge 0$ while $5-x$ can be any real number. But if you add the restriction $5-x ge 0$, then $25 geq |5 - x|$ becomes

$$text$25 geq 5 - x textand 5-x ge 0$$$

share|cite|improve this answer

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

$endgroup$

share|cite|improve this answer

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

answered Apr 27 at 17:08

steven gregorysteven gregory

18.6k32459

1

$begingroup$

If you look carefully, you'll notice your definition has the square inside
the square root (not outside):
$$|a| = sqrta^2$$
However, in your solution you seem to assume that:
$$sqrta^2 = (sqrta)^2$$
However,
$$sqrta^2 neq (sqrta)^2$$
For example, as others have suggested, if $a = -1$
we have:
$$sqrta^2 = sqrt(-1)^2 = sqrt1 = 1$$
but
$$(sqrta)^2 = (sqrt-1)^2 = i^2 = -1$$

share|cite|improve this answer

answered 2 days ago

user669329user669329

111

New contributor

user669329 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

If you look carefully, you'll notice your definition has the square inside
the square root (not outside):
$$|a| = sqrta^2$$
However, in your solution you seem to assume that:
$$sqrta^2 = (sqrta)^2$$
However,
$$sqrta^2 neq (sqrta)^2$$
For example, as others have suggested, if $a = -1$
we have:
$$sqrta^2 = sqrt(-1)^2 = sqrt1 = 1$$
but
$$(sqrta)^2 = (sqrt-1)^2 = i^2 = -1$$

share|cite|improve this answer

answered 2 days ago

user669329user669329

111

New contributor

user669329 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

If you look carefully, you'll notice your definition has the square inside
the square root (not outside):
$$|a| = sqrta^2$$
However, in your solution you seem to assume that:
$$sqrta^2 = (sqrta)^2$$
However,
$$sqrta^2 neq (sqrta)^2$$
For example, as others have suggested, if $a = -1$
we have:
$$sqrta^2 = sqrt(-1)^2 = sqrt1 = 1$$
but
$$(sqrta)^2 = (sqrt-1)^2 = i^2 = -1$$

share|cite|improve this answer

answered 2 days ago

user669329user669329

111

New contributor

user669329 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 answer

answered 2 days ago

user669329user669329

111

New contributor

user669329 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

user669329user669329

111

New contributor

user669329 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

user669329user669329

111