theory of equation..finding roots of a modular function

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The number of real roots of $big|x^2+4|x|+3big| +2x-11=0$ is?
I have tried by expanding modulus, and if I'm right I'm getting two real roots but I'm a bit confused because of the innermost modulus, please help

edited Jul 23 at 3:41

dxiv

54k64796

asked Jul 23 at 3:35

Nitin Jha

1

I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are.
â€“Â dxiv
Jul 23 at 3:39

Are we consider $x in mathbbR$ or $x in mathbbC$? I presume the former, but just checking
â€“Â Brevan Ellefsen
Jul 23 at 3:51

add a commentÂ |Â

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edited Jul 23 at 3:41

dxiv

54k64796

asked Jul 23 at 3:35

Nitin Jha

1

I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are.
â€“Â dxiv
Jul 23 at 3:39

Are we consider $x in mathbbR$ or $x in mathbbC$? I presume the former, but just checking
â€“Â Brevan Ellefsen
Jul 23 at 3:51

add a commentÂ |Â

up vote
0
down vote

favorite

edited Jul 23 at 3:41

dxiv

54k64796

asked Jul 23 at 3:35

Nitin Jha

edited Jul 23 at 3:41

dxiv

54k64796

asked Jul 23 at 3:35

Nitin Jha

edited Jul 23 at 3:41

dxiv

54k64796

edited Jul 23 at 3:41

dxiv

54k64796

edited Jul 23 at 3:41

dxiv

54k64796

asked Jul 23 at 3:35

Nitin Jha

asked Jul 23 at 3:35

Nitin Jha

asked Jul 23 at 3:35

Nitin Jha

1

I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are.
â€“Â dxiv
Jul 23 at 3:39

Are we consider $x in mathbbR$ or $x in mathbbC$? I presume the former, but just checking
â€“Â Brevan Ellefsen
Jul 23 at 3:51

add a commentÂ |Â

1

I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are.
â€“Â dxiv
Jul 23 at 3:39

Are we consider $x in mathbbR$ or $x in mathbbC$? I presume the former, but just checking
â€“Â Brevan Ellefsen
Jul 23 at 3:51

I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are.
â€“Â dxiv
Jul 23 at 3:39

Are we consider $x in mathbbR$ or $x in mathbbC$? I presume the former, but just checking
â€“Â Brevan Ellefsen
Jul 23 at 3:51

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

If $xge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = sqrt17-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

If $2-sqrt15 le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)

If $x < 2-sqrt15$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-sqrt15$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = sqrt17-3$ and $x = -2 qquad square$

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

add a commentÂ |Â

up vote
0
down vote

First answer this question: Under what circumstances can
$x^2 + 4lvert xrvert + 3$ be negative?

This will simplify the problem quite a bit.

Remember that a formula like this can be expressed as a piecewise polynomial function if you can figure out the values of $x$ at which anything inside an absolute value sign goes from negative to positive or vice versa.
Within any interval where you do not have such a sign change, the function is polynomial.
In this problem there are fewer pieces than you might expect.

answered Jul 23 at 3:54

David K

48.2k340107

add a commentÂ |Â

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$$big|x^2+4|x|+3big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3pm sqrt 17$

We keep the positive one.
$ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$big|x^2-4x+3big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+sqrt 17$

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

If $xge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = sqrt17-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

If $2-sqrt15 le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)

If $x < 2-sqrt15$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-sqrt15$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = sqrt17-3$ and $x = -2 qquad square$

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

add a commentÂ |Â

up vote
1
down vote

If $xge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = sqrt17-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

If $2-sqrt15 le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)

If $x < 2-sqrt15$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-sqrt15$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = sqrt17-3$ and $x = -2 qquad square$

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

add a commentÂ |Â

up vote
1
down vote

If $xge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = sqrt17-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

If $2-sqrt15 le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)

If $x < 2-sqrt15$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-sqrt15$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = sqrt17-3$ and $x = -2 qquad square$

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

If $xge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = sqrt17-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

If $2-sqrt15 le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)

If $x < 2-sqrt15$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-sqrt15$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = sqrt17-3$ and $x = -2 qquad square$

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

edited Jul 23 at 4:23

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

answered Jul 23 at 4:11

Brevan Ellefsen

11.4k31449

add a commentÂ |Â

up vote
0
down vote

First answer this question: Under what circumstances can
$x^2 + 4lvert xrvert + 3$ be negative?

This will simplify the problem quite a bit.

answered Jul 23 at 3:54

David K

48.2k340107

add a commentÂ |Â

up vote
0
down vote

First answer this question: Under what circumstances can
$x^2 + 4lvert xrvert + 3$ be negative?

This will simplify the problem quite a bit.

answered Jul 23 at 3:54

David K

48.2k340107

add a commentÂ |Â

up vote
0
down vote

First answer this question: Under what circumstances can
$x^2 + 4lvert xrvert + 3$ be negative?

This will simplify the problem quite a bit.

answered Jul 23 at 3:54

David K

48.2k340107

First answer this question: Under what circumstances can
$x^2 + 4lvert xrvert + 3$ be negative?

This will simplify the problem quite a bit.

answered Jul 23 at 3:54

David K

48.2k340107

answered Jul 23 at 3:54

David K

48.2k340107

answered Jul 23 at 3:54

David K

48.2k340107

answered Jul 23 at 3:54

David K

48.2k340107

add a commentÂ |Â

up vote
0
down vote

$$big|x^2+4|x|+3big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3pm sqrt 17$

We keep the positive one.
$ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$big|x^2-4x+3big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+sqrt 17$

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

add a commentÂ |Â

up vote
0
down vote

$$big|x^2+4|x|+3big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3pm sqrt 17$

We keep the positive one.
$ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$big|x^2-4x+3big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+sqrt 17$

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

add a commentÂ |Â

up vote
0
down vote

$$big|x^2+4|x|+3big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3pm sqrt 17$

We keep the positive one.
$ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$big|x^2-4x+3big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+sqrt 17$

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

$$big|x^2+4|x|+3big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3pm sqrt 17$

We keep the positive one.
$ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$big|x^2-4x+3big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+sqrt 17$

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

edited Jul 23 at 4:34

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

answered Jul 23 at 4:17

Mohammad Riazi-Kermani

27.5k41852

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

add a commentÂ |Â

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

Two of those "solutions" aren't even solutions: $|(2)^2+4|(2)|+3big| +2(2)-11= 19-11=8 neq0$ $|(4)^2+4|(4)|+3big| +2(4)-11= 32 neq0$ Moreover, you missed the solution $sqrt17-3$, which works since $|(sqrt17-3)^2+4|(sqrt17-3)|+3big| +2(sqrt17-3)-11 = (17 - 2sqrt17)+2(sqrt17-3)-11 = 0$
â€“Â Brevan Ellefsen
Jul 23 at 4:26

@BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills.
â€“Â Mohammad Riazi-Kermani
Jul 23 at 4:38

add a commentÂ |Â

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