Absolute value with factors in it [closed]
$begingroup$
$$ lvert1+x(1-x)rvert< lvert1-xrvert$$
How do I solve for $x$? I'm having a hard time finding the intervals.
inequality absolute-value
edited Jan 22 at 23:21
Joseph Eck
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
$endgroup$
closed as off-topic by Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen Jan 23 at 5:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$ lvert1+x(1-x)rvert< lvert1-xrvert$$
How do I solve for $x$? I'm having a hard time finding the intervals.
inequality absolute-value
edited Jan 22 at 23:21
Joseph Eck
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
$endgroup$
closed as off-topic by Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen Jan 23 at 5:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$ lvert1+x(1-x)rvert< lvert1-xrvert$$
How do I solve for $x$? I'm having a hard time finding the intervals.
inequality absolute-value
edited Jan 22 at 23:21
Joseph Eck
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
$endgroup$
$$ lvert1+x(1-x)rvert< lvert1-xrvert$$
How do I solve for $x$? I'm having a hard time finding the intervals.
inequality absolute-value
inequality absolute-value
edited Jan 22 at 23:21
Joseph Eck
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
edited Jan 22 at 23:21
Joseph Eck
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
edited Jan 22 at 23:21
Joseph Eck
563413
edited Jan 22 at 23:21
Joseph Eck
563413
edited Jan 22 at 23:21
Joseph Eck
563413
563413
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
asked Jan 22 at 23:07
BullerskyddBullerskydd
73
73
closed as off-topic by Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen Jan 23 at 5:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen Jan 23 at 5:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, Shaun, RRL, max_zorn, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$lvert1+x(1-x)rvert< lvert1-xrvert$
$Leftrightarrow |-(x^2-x-1)|<|1-x|$
$Leftrightarrow (x^2-x-1)^2<(1-x)^2$
$Leftrightarrow x(x-2)(x^2-2)<0$
Can you solve it now?
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
$endgroup$
add a comment |
$begingroup$
If $x geq 1$ then the inequality implies $1+x(1-x) < x-1$ or which give $x >sqrt 2$. But we also need $-1-x(1-x) < x-1$ which means $x <2$. Hence $sqrt 2 <x<2$. Similarly, for $x<1$ we get $-sqrt 2 <x<0$. Hence the solution set is $(-sqrt 2,0)cup (sqrt 2 ,2)$.
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$lvert1+x(1-x)rvert< lvert1-xrvert$
$Leftrightarrow |-(x^2-x-1)|<|1-x|$
$Leftrightarrow (x^2-x-1)^2<(1-x)^2$
$Leftrightarrow x(x-2)(x^2-2)<0$
Can you solve it now?
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
$endgroup$
add a comment |
$begingroup$
$lvert1+x(1-x)rvert< lvert1-xrvert$
$Leftrightarrow |-(x^2-x-1)|<|1-x|$
$Leftrightarrow (x^2-x-1)^2<(1-x)^2$
$Leftrightarrow x(x-2)(x^2-2)<0$
Can you solve it now?
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
$endgroup$
add a comment |
$begingroup$
$lvert1+x(1-x)rvert< lvert1-xrvert$
$Leftrightarrow |-(x^2-x-1)|<|1-x|$
$Leftrightarrow (x^2-x-1)^2<(1-x)^2$
$Leftrightarrow x(x-2)(x^2-2)<0$
Can you solve it now?
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
$endgroup$
$lvert1+x(1-x)rvert< lvert1-xrvert$
$Leftrightarrow |-(x^2-x-1)|<|1-x|$
$Leftrightarrow (x^2-x-1)^2<(1-x)^2$
$Leftrightarrow x(x-2)(x^2-2)<0$
Can you solve it now?
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
answered Jan 22 at 23:44
Word ShallowWord Shallow
8241618
8241618
add a comment |
add a comment |
$begingroup$
If $x geq 1$ then the inequality implies $1+x(1-x) < x-1$ or which give $x >sqrt 2$. But we also need $-1-x(1-x) < x-1$ which means $x <2$. Hence $sqrt 2 <x<2$. Similarly, for $x<1$ we get $-sqrt 2 <x<0$. Hence the solution set is $(-sqrt 2,0)cup (sqrt 2 ,2)$.
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
If $x geq 1$ then the inequality implies $1+x(1-x) < x-1$ or which give $x >sqrt 2$. But we also need $-1-x(1-x) < x-1$ which means $x <2$. Hence $sqrt 2 <x<2$. Similarly, for $x<1$ we get $-sqrt 2 <x<0$. Hence the solution set is $(-sqrt 2,0)cup (sqrt 2 ,2)$.
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
If $x geq 1$ then the inequality implies $1+x(1-x) < x-1$ or which give $x >sqrt 2$. But we also need $-1-x(1-x) < x-1$ which means $x <2$. Hence $sqrt 2 <x<2$. Similarly, for $x<1$ we get $-sqrt 2 <x<0$. Hence the solution set is $(-sqrt 2,0)cup (sqrt 2 ,2)$.
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
If $x geq 1$ then the inequality implies $1+x(1-x) < x-1$ or which give $x >sqrt 2$. But we also need $-1-x(1-x) < x-1$ which means $x <2$. Hence $sqrt 2 <x<2$. Similarly, for $x<1$ we get $-sqrt 2 <x<0$. Hence the solution set is $(-sqrt 2,0)cup (sqrt 2 ,2)$.
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 22 at 23:41
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
add a comment |
add a comment |
