Mixing
Find parameters that satisfy two conditions of an equation [duplicate]
$begingroup$
This question already has an answer here:
Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition.
2 answers
There is the following equation.
$$x^{2}+2(m-a)x+3am-2=0$$
a) Find $a$ such that the equation has real roots, $forall min mathbb{R}$
b) Find $m$ such that the equation has real roots, $forall ain mathbb{R}$
The discriminant is $m^{2}-5am+a^{2}+2geq 0$ and $a^{2}-5am+m^{2}+2geq 0$
I found that $a,min (-infty , -sqrt{frac{8}{21}}]cup [sqrt{frac{8}{21}}, +infty)$ but I don't know why $|a|leqslant sqrt{frac{8}{21}}$and $|m|leqslant sqrt{frac{8}{21}}$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $le0$. For radicals I know that it should be $ge0$.
linear-algebra algebra-precalculus polynomials
edited Jan 18 at 19:27
greedoid
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
$endgroup$
marked as duplicate by Dietrich Burde, greedoid, Adrian Keister, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jan 19 at 18:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition.
2 answers
There is the following equation.
$$x^{2}+2(m-a)x+3am-2=0$$
a) Find $a$ such that the equation has real roots, $forall min mathbb{R}$
b) Find $m$ such that the equation has real roots, $forall ain mathbb{R}$
The discriminant is $m^{2}-5am+a^{2}+2geq 0$ and $a^{2}-5am+m^{2}+2geq 0$
I found that $a,min (-infty , -sqrt{frac{8}{21}}]cup [sqrt{frac{8}{21}}, +infty)$ but I don't know why $|a|leqslant sqrt{frac{8}{21}}$and $|m|leqslant sqrt{frac{8}{21}}$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $le0$. For radicals I know that it should be $ge0$.
linear-algebra algebra-precalculus polynomials
edited Jan 18 at 19:27
greedoid
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
$endgroup$
marked as duplicate by Dietrich Burde, greedoid, Adrian Keister, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jan 19 at 18:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07
add a comment |
$begingroup$
This question already has an answer here:
Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition.
2 answers
There is the following equation.
$$x^{2}+2(m-a)x+3am-2=0$$
a) Find $a$ such that the equation has real roots, $forall min mathbb{R}$
b) Find $m$ such that the equation has real roots, $forall ain mathbb{R}$
The discriminant is $m^{2}-5am+a^{2}+2geq 0$ and $a^{2}-5am+m^{2}+2geq 0$
I found that $a,min (-infty , -sqrt{frac{8}{21}}]cup [sqrt{frac{8}{21}}, +infty)$ but I don't know why $|a|leqslant sqrt{frac{8}{21}}$and $|m|leqslant sqrt{frac{8}{21}}$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $le0$. For radicals I know that it should be $ge0$.
linear-algebra algebra-precalculus polynomials
edited Jan 18 at 19:27
greedoid
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
$endgroup$
This question already has an answer here:
Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition.
2 answers
There is the following equation.
$$x^{2}+2(m-a)x+3am-2=0$$
a) Find $a$ such that the equation has real roots, $forall min mathbb{R}$
b) Find $m$ such that the equation has real roots, $forall ain mathbb{R}$
The discriminant is $m^{2}-5am+a^{2}+2geq 0$ and $a^{2}-5am+m^{2}+2geq 0$
I found that $a,min (-infty , -sqrt{frac{8}{21}}]cup [sqrt{frac{8}{21}}, +infty)$ but I don't know why $|a|leqslant sqrt{frac{8}{21}}$and $|m|leqslant sqrt{frac{8}{21}}$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $le0$. For radicals I know that it should be $ge0$.
This question already has an answer here:
Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition.
2 answers
linear-algebra algebra-precalculus polynomials
linear-algebra algebra-precalculus polynomials
edited Jan 18 at 19:27
greedoid
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
edited Jan 18 at 19:27
greedoid
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
edited Jan 18 at 19:27
greedoid
44.8k1156111
edited Jan 18 at 19:27
greedoid
44.8k1156111
edited Jan 18 at 19:27
greedoid
44.8k1156111
44.8k1156111
asked Jan 18 at 18:48
Vali ROVali RO
686
asked Jan 18 at 18:48
Vali ROVali RO
686
asked Jan 18 at 18:48
Vali ROVali RO
686
686
marked as duplicate by Dietrich Burde, greedoid, Adrian Keister, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jan 19 at 18:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, greedoid, Adrian Keister, José Carlos Santos linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jan 19 at 18:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07
add a comment |
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So you have a discriminant $$Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$Delta_2(a)=25 a^2-4(a^2+2)<0$$
This will ensure that $Delta_1(m,a)>0$ for any $m$.
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So you have a discriminant $$Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$Delta_2(a)=25 a^2-4(a^2+2)<0$$
This will ensure that $Delta_1(m,a)>0$ for any $m$.
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
add a comment |
$begingroup$
So you have a discriminant $$Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$Delta_2(a)=25 a^2-4(a^2+2)<0$$
This will ensure that $Delta_1(m,a)>0$ for any $m$.
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
add a comment |
$begingroup$
So you have a discriminant $$Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$Delta_2(a)=25 a^2-4(a^2+2)<0$$
This will ensure that $Delta_1(m,a)>0$ for any $m$.
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
$endgroup$
So you have a discriminant $$Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$Delta_2(a)=25 a^2-4(a^2+2)<0$$
This will ensure that $Delta_1(m,a)>0$ for any $m$.
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
answered Jan 18 at 19:08
AndreiAndrei
12.4k21128
12.4k21128
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
add a comment |
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex.
$endgroup$
– Vali RO
Jan 18 at 19:27
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below.
$endgroup$
– Andrei
Jan 18 at 19:36
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
$begingroup$
I finally understood.Thank you so much for your help!
$endgroup$
– Vali RO
Jan 18 at 19:41
add a comment |
$begingroup$
Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0
$endgroup$
– Vali RO
Jan 18 at 18:56
$begingroup$
You could ask this first at the original question before posting a duplicate.
$endgroup$
– Dietrich Burde
Jan 18 at 18:59
$begingroup$
$|a|leq sqrt{frac{8}{21}}$ because for $m^2-5am + a^2 +2 geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$.
$endgroup$
– Sauhard Sharma
Jan 18 at 19:07