An inequality question with three unknowns The 2019 Stack Overflow Developer Survey Results Are InSolve three simultaneous equations with three unknownsAbsolute inequality question.The solution set of an inequalitySolve three equations for three unknowns.Finding unknowns from three equal equationsSolve three equations with three unknownsAn algebraic inequality with three componentsInequality exercise with greater integerInequality exercise with floor functionHow do I find the solution set to the inequality $3x^2-4x+5<0$
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An inequality question with three unknowns
The 2019 Stack Overflow Developer Survey Results Are InSolve three simultaneous equations with three unknownsAbsolute inequality question.The solution set of an inequalitySolve three equations for three unknowns.Finding unknowns from three equal equationsSolve three equations with three unknownsAn algebraic inequality with three componentsInequality exercise with greater integerInequality exercise with floor functionHow do I find the solution set to the inequality $3x^2-4x+5<0$
$begingroup$
$x^2+mx+2n-1>0$
If the solution set of the inequality is $R-(5)$. What is the sum of $m+n$?
algebra-precalculus
asked Mar 23 at 16:54
Hatm02Hatm02
103
$endgroup$
add a comment |
$begingroup$
$x^2+mx+2n-1>0$
If the solution set of the inequality is $R-(5)$. What is the sum of $m+n$?
algebra-precalculus
asked Mar 23 at 16:54
Hatm02Hatm02
103
$endgroup$
add a comment |
$begingroup$
$x^2+mx+2n-1>0$
If the solution set of the inequality is $R-(5)$. What is the sum of $m+n$?
algebra-precalculus
asked Mar 23 at 16:54
Hatm02Hatm02
103
$endgroup$
$x^2+mx+2n-1>0$
If the solution set of the inequality is $R-(5)$. What is the sum of $m+n$?
algebra-precalculus
algebra-precalculus
asked Mar 23 at 16:54
Hatm02Hatm02
103
asked Mar 23 at 16:54
Hatm02Hatm02
103
asked Mar 23 at 16:54
Hatm02Hatm02
103
asked Mar 23 at 16:54
Hatm02Hatm02
103
asked Mar 23 at 16:54
Hatm02Hatm02
103
103
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:Write your inequality in the form $$left(x+fracm2right)^2+2n-1-fracm^24>0$$
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
add a comment |
$begingroup$
HINT:$$left(x+fracm2right)^2+2n-1-fracm^24>0$$ after that what i could think of is that x=5 is the point where $$left(x+fracm2right)^2$$ turns zero and $$2n-1-fracm^24$$ turns zero simultaneously..
answered Mar 25 at 19:12
user224359user224359
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:Write your inequality in the form $$left(x+fracm2right)^2+2n-1-fracm^24>0$$
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
add a comment |
$begingroup$
Hint:Write your inequality in the form $$left(x+fracm2right)^2+2n-1-fracm^24>0$$
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
add a comment |
$begingroup$
Hint:Write your inequality in the form $$left(x+fracm2right)^2+2n-1-fracm^24>0$$
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
Hint:Write your inequality in the form $$left(x+fracm2right)^2+2n-1-fracm^24>0$$
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Mar 23 at 17:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
78.9k42867
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
add a comment |
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
I wrote it in that form but I couldn't find the solution.
$endgroup$
– Hatm02
Mar 23 at 17:05
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
should your inequality be fulfilled for all real $x$?
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:07
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Yes, you're right.
$endgroup$
– Hatm02
Mar 23 at 17:10
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
$begingroup$
Then it must be $$2n-1-fracm^24=0$$ and $$left(x+fracm2right)^2neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Mar 23 at 17:12
add a comment |
$begingroup$
HINT:$$left(x+fracm2right)^2+2n-1-fracm^24>0$$ after that what i could think of is that x=5 is the point where $$left(x+fracm2right)^2$$ turns zero and $$2n-1-fracm^24$$ turns zero simultaneously..
answered Mar 25 at 19:12
user224359user224359
1
$endgroup$
add a comment |
$begingroup$
HINT:$$left(x+fracm2right)^2+2n-1-fracm^24>0$$ after that what i could think of is that x=5 is the point where $$left(x+fracm2right)^2$$ turns zero and $$2n-1-fracm^24$$ turns zero simultaneously..
answered Mar 25 at 19:12
user224359user224359
1
$endgroup$
add a comment |
$begingroup$
HINT:$$left(x+fracm2right)^2+2n-1-fracm^24>0$$ after that what i could think of is that x=5 is the point where $$left(x+fracm2right)^2$$ turns zero and $$2n-1-fracm^24$$ turns zero simultaneously..
answered Mar 25 at 19:12
user224359user224359
1
$endgroup$
HINT:$$left(x+fracm2right)^2+2n-1-fracm^24>0$$ after that what i could think of is that x=5 is the point where $$left(x+fracm2right)^2$$ turns zero and $$2n-1-fracm^24$$ turns zero simultaneously..
answered Mar 25 at 19:12
user224359user224359
1
answered Mar 25 at 19:12
user224359user224359
1
answered Mar 25 at 19:12
user224359user224359
1
answered Mar 25 at 19:12
user224359user224359
1
1
add a comment |
add a comment |
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