Find the values of $a,b in Bbb R$ (if exists) such that $-5 le fracx^2+ax+bx^2+2x+3 le 4$ for all $x in Bbb R$ Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Range of values of f(x) using quadratic inequalities (need intuition)Find all values of parameter a, when sum of solutions of following equation is 100Find the values of $b$ for which the equation $2log_frac125(bx+28)=-log_5(12-4x-x^2)$ has only one solutionFind the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions?Find the interval of $c$ such that the rational function $fracx^2+2x+cx^2+4x+3c$ takes all real valuesFind the values of $c$ for which the vectors make an obtuse angle for any $xepsilon(0,infty)$Find the domain of the inequation $x^2 -ax +1-2a^2 > 0$ for all $xin Bbb R$Finding roots of a quadratic equation having same roots for all real values of a parameterFind all values of $k$ for $kx^2+(k+2)x-3=0$ with positive roots.Find values of $a$ such that $x^2+ax+a^2+6a lt 0$ $forall$ $x in (1,2)$
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Find the values of $a,b in Bbb R$ (if exists) such that $-5 le fracx^2+ax+bx^2+2x+3 le 4$ for all $x in Bbb R$
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Range of values of f(x) using quadratic inequalities (need intuition)Find all values of parameter a, when sum of solutions of following equation is 100Find the values of $b$ for which the equation $2log_frac125(bx+28)=-log_5(12-4x-x^2)$ has only one solutionFind the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions?Find the interval of $c$ such that the rational function $fracx^2+2x+cx^2+4x+3c$ takes all real valuesFind the values of $c$ for which the vectors make an obtuse angle for any $xepsilon(0,infty)$Find the domain of the inequation $x^2 -ax +1-2a^2 > 0$ for all $xin Bbb R$Finding roots of a quadratic equation having same roots for all real values of a parameterFind all values of $k$ for $kx^2+(k+2)x-3=0$ with positive roots.Find values of $a$ such that $x^2+ax+a^2+6a lt 0$ $forall$ $x in (1,2)$
$begingroup$
Find the values of $a,b in Bbb R$ (if exists) such that $$-5 le fracx^2+ax+bx^2+2x+3 le 4$$ for all $x in Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)le x^2+ax+b $ $space$ $land$ $space$ $4(x^2+2x+3) ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
inequality quadratics
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
$endgroup$
add a comment |
$begingroup$
Find the values of $a,b in Bbb R$ (if exists) such that $$-5 le fracx^2+ax+bx^2+2x+3 le 4$$ for all $x in Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)le x^2+ax+b $ $space$ $land$ $space$ $4(x^2+2x+3) ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
inequality quadratics
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
$endgroup$
1
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54
add a comment |
$begingroup$
Find the values of $a,b in Bbb R$ (if exists) such that $$-5 le fracx^2+ax+bx^2+2x+3 le 4$$ for all $x in Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)le x^2+ax+b $ $space$ $land$ $space$ $4(x^2+2x+3) ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
inequality quadratics
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
$endgroup$
Find the values of $a,b in Bbb R$ (if exists) such that $$-5 le fracx^2+ax+bx^2+2x+3 le 4$$ for all $x in Bbb R$
My try:
I noticed that $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)le x^2+ax+b $ $space$ $land$ $space$ $4(x^2+2x+3) ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints?
inequality quadratics
inequality quadratics
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
edited Mar 31 at 22:53
Rodrigo Pizarro
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
asked Mar 31 at 22:46
Rodrigo PizarroRodrigo Pizarro
1,034219
1,034219
1
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54
add a comment |
1
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54
1
1
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $a,binBbbR$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)geq0,$$
$$3x^2+(8-a)x+(12-b)geq0,$$
for all $xinBbbR$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)leq0,$$
$$(8-a)^2-12(12-b)leq0.$$
Isolating $b$ from both inequalities yields
$$frac(a+10)^224-15leq bleq 12-frac(8-a)^212.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|leq4sqrttfrac343.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
$endgroup$
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
add a comment |
$begingroup$
Hint: $$x^2+2x+3=(x+1)^2+2ge2$$
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Let $a,binBbbR$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)geq0,$$
$$3x^2+(8-a)x+(12-b)geq0,$$
for all $xinBbbR$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)leq0,$$
$$(8-a)^2-12(12-b)leq0.$$
Isolating $b$ from both inequalities yields
$$frac(a+10)^224-15leq bleq 12-frac(8-a)^212.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|leq4sqrttfrac343.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
$endgroup$
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
add a comment |
$begingroup$
Let $a,binBbbR$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)geq0,$$
$$3x^2+(8-a)x+(12-b)geq0,$$
for all $xinBbbR$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)leq0,$$
$$(8-a)^2-12(12-b)leq0.$$
Isolating $b$ from both inequalities yields
$$frac(a+10)^224-15leq bleq 12-frac(8-a)^212.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|leq4sqrttfrac343.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
$endgroup$
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
add a comment |
$begingroup$
Let $a,binBbbR$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)geq0,$$
$$3x^2+(8-a)x+(12-b)geq0,$$
for all $xinBbbR$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)leq0,$$
$$(8-a)^2-12(12-b)leq0.$$
Isolating $b$ from both inequalities yields
$$frac(a+10)^224-15leq bleq 12-frac(8-a)^212.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|leq4sqrttfrac343.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
$endgroup$
Let $a,binBbbR$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities
$$6x^2+(a+10)x+(b+15)geq0,$$
$$3x^2+(8-a)x+(12-b)geq0,$$
for all $xinBbbR$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that
$$(a+10)^2-24(b+15)leq0,$$
$$(8-a)^2-12(12-b)leq0.$$
Isolating $b$ from both inequalities yields
$$frac(a+10)^224-15leq bleq 12-frac(8-a)^212.$$
Moreover, a bit of algebra shows that for $a$ we then have
$$3a^2-12a-532leq0.$$
By the quadratic formula this is equivalent to
$$|a-2|leq4sqrttfrac343.$$
Can you finish from here?
For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
edited Mar 31 at 23:52
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
answered Mar 31 at 22:51
ServaesServaes
30.6k342101
30.6k342101
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
add a comment |
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
$begingroup$
I exactly did that, but i don't know how to proceed with the discriminants.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 23:28
1
1
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
$begingroup$
I have added a few more steps, but I have left the end of the proof for you.
$endgroup$
– Servaes
Mar 31 at 23:32
add a comment |
$begingroup$
Hint: $$x^2+2x+3=(x+1)^2+2ge2$$
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
$endgroup$
add a comment |
$begingroup$
Hint: $$x^2+2x+3=(x+1)^2+2ge2$$
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
$endgroup$
add a comment |
$begingroup$
Hint: $$x^2+2x+3=(x+1)^2+2ge2$$
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
$endgroup$
Hint: $$x^2+2x+3=(x+1)^2+2ge2$$
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
answered Mar 31 at 22:50
Dr. MathvaDr. Mathva
3,5441630
3,5441630
add a comment |
add a comment |
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1
$begingroup$
Do you mean that the inequalities should hold for all $xinBbbR$?
$endgroup$
– Servaes
Mar 31 at 22:50
$begingroup$
Yes, i edited the question.
$endgroup$
– Rodrigo Pizarro
Mar 31 at 22:54