If $frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$ then $(a-b)^3 le0$ [closed]
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0 implies (a-b)^3 le0$$
Can someone explain why this is true? I have tried doing it by expanding all the terms but can seem to prove the identity
algebra-precalculus inequality
edited Dec 10 '18 at 20:39
Did
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
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closed as off-topic by Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz Dec 10 '18 at 15:01
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." – Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0 implies (a-b)^3 le0$$
Can someone explain why this is true? I have tried doing it by expanding all the terms but can seem to prove the identity
algebra-precalculus inequality
edited Dec 10 '18 at 20:39
Did
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
$endgroup$
closed as off-topic by Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz Dec 10 '18 at 15:01
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." – Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Note there is the obvious factor $a-b$ in the LHS, so ...
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– user10354138
Dec 9 '18 at 17:49
add a comment |
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0 implies (a-b)^3 le0$$
Can someone explain why this is true? I have tried doing it by expanding all the terms but can seem to prove the identity
algebra-precalculus inequality
edited Dec 10 '18 at 20:39
Did
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
$endgroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0 implies (a-b)^3 le0$$
Can someone explain why this is true? I have tried doing it by expanding all the terms but can seem to prove the identity
algebra-precalculus inequality
algebra-precalculus inequality
edited Dec 10 '18 at 20:39
Did
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
edited Dec 10 '18 at 20:39
Did
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
edited Dec 10 '18 at 20:39
Did
247k23222458
edited Dec 10 '18 at 20:39
Did
247k23222458
edited Dec 10 '18 at 20:39
Did
247k23222458
247k23222458
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
asked Dec 9 '18 at 17:44
pablo_mathscobarpablo_mathscobar
906
906
closed as off-topic by Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz Dec 10 '18 at 15:01
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." – Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz Dec 10 '18 at 15:01
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." – Nosrati, José Carlos Santos, John B, Saad, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Note there is the obvious factor $a-b$ in the LHS, so ...
$endgroup$
– user10354138
Dec 9 '18 at 17:49
add a comment |
$begingroup$
Note there is the obvious factor $a-b$ in the LHS, so ...
$endgroup$
– user10354138
Dec 9 '18 at 17:49
$begingroup$
Note there is the obvious factor $a-b$ in the LHS, so ...
$endgroup$
– user10354138
Dec 9 '18 at 17:49
$begingroup$
Note there is the obvious factor $a-b$ in the LHS, so ...
$endgroup$
– user10354138
Dec 9 '18 at 17:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is enought to prove $aleq b$. So say $a>b$, then from
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$$
we get $$frac{1}{4}(b^2-a^2) + frac{1}{3}(a^2+ab + b^2)le0$$
so $$3(b^2-a^2) + 4(a^2+ab + b^2)le0$$
so $$3b^2 +(2b+a)^2 = 7b^2+4ab+a^2 leq 0$$
a contradiction (since it is true only if $b=0$ and $2b+a=0$).
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
$endgroup$
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
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– pablo_mathscobar
Dec 9 '18 at 20:06
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Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
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– greedoid
Dec 9 '18 at 20:20
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cheers mate, appreciate the help
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– pablo_mathscobar
Dec 9 '18 at 20:41
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Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
add a comment |
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3){={1over 12}(a-b)Bigg(7b^2+a^2+4abBigg)}$$also $7b^2+a^2+4ab$ is always non-negative (as a function of $b$ and for different values of $a$ where $7b^2+a^2+4ab=0$ iff $a=b=0$) since for $ane 0$ we have $Delta<0$ ($Delta={4^2a^2-4times 7a^2}=-12a^2<0$) . therefore $$a-ble 0$$and $$(a-b)^3le 0$$
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
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I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
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i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
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but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 19:43
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is enought to prove $aleq b$. So say $a>b$, then from
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$$
we get $$frac{1}{4}(b^2-a^2) + frac{1}{3}(a^2+ab + b^2)le0$$
so $$3(b^2-a^2) + 4(a^2+ab + b^2)le0$$
so $$3b^2 +(2b+a)^2 = 7b^2+4ab+a^2 leq 0$$
a contradiction (since it is true only if $b=0$ and $2b+a=0$).
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
$endgroup$
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
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cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
add a comment |
$begingroup$
It is enought to prove $aleq b$. So say $a>b$, then from
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$$
we get $$frac{1}{4}(b^2-a^2) + frac{1}{3}(a^2+ab + b^2)le0$$
so $$3(b^2-a^2) + 4(a^2+ab + b^2)le0$$
so $$3b^2 +(2b+a)^2 = 7b^2+4ab+a^2 leq 0$$
a contradiction (since it is true only if $b=0$ and $2b+a=0$).
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
$endgroup$
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
$begingroup$
cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
add a comment |
$begingroup$
It is enought to prove $aleq b$. So say $a>b$, then from
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$$
we get $$frac{1}{4}(b^2-a^2) + frac{1}{3}(a^2+ab + b^2)le0$$
so $$3(b^2-a^2) + 4(a^2+ab + b^2)le0$$
so $$3b^2 +(2b+a)^2 = 7b^2+4ab+a^2 leq 0$$
a contradiction (since it is true only if $b=0$ and $2b+a=0$).
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
$endgroup$
It is enought to prove $aleq b$. So say $a>b$, then from
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3)le0$$
we get $$frac{1}{4}(b^2-a^2) + frac{1}{3}(a^2+ab + b^2)le0$$
so $$3(b^2-a^2) + 4(a^2+ab + b^2)le0$$
so $$3b^2 +(2b+a)^2 = 7b^2+4ab+a^2 leq 0$$
a contradiction (since it is true only if $b=0$ and $2b+a=0$).
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
answered Dec 9 '18 at 17:51
greedoidgreedoid
39.2k114797
39.2k114797
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
$begingroup$
cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
add a comment |
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
$begingroup$
cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
why is the last line a contradiction? i cant see why 7b^2+4ab+a^2 is >0
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:06
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
$begingroup$
Because from $2b+a=0$ and $b=0$ we get $a=0$, but $a>b$
$endgroup$
– greedoid
Dec 9 '18 at 20:20
$begingroup$
cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
cheers mate, appreciate the help
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 20:41
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
$begingroup$
Glad to help, don't forget to upvote and accept if you like.
$endgroup$
– greedoid
Dec 9 '18 at 20:45
add a comment |
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3){={1over 12}(a-b)Bigg(7b^2+a^2+4abBigg)}$$also $7b^2+a^2+4ab$ is always non-negative (as a function of $b$ and for different values of $a$ where $7b^2+a^2+4ab=0$ iff $a=b=0$) since for $ane 0$ we have $Delta<0$ ($Delta={4^2a^2-4times 7a^2}=-12a^2<0$) . therefore $$a-ble 0$$and $$(a-b)^3le 0$$
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
$begingroup$
but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 19:43
|
show 2 more comments
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3){={1over 12}(a-b)Bigg(7b^2+a^2+4abBigg)}$$also $7b^2+a^2+4ab$ is always non-negative (as a function of $b$ and for different values of $a$ where $7b^2+a^2+4ab=0$ iff $a=b=0$) since for $ane 0$ we have $Delta<0$ ($Delta={4^2a^2-4times 7a^2}=-12a^2<0$) . therefore $$a-ble 0$$and $$(a-b)^3le 0$$
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
$begingroup$
but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 19:43
|
show 2 more comments
$begingroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3){={1over 12}(a-b)Bigg(7b^2+a^2+4abBigg)}$$also $7b^2+a^2+4ab$ is always non-negative (as a function of $b$ and for different values of $a$ where $7b^2+a^2+4ab=0$ iff $a=b=0$) since for $ane 0$ we have $Delta<0$ ($Delta={4^2a^2-4times 7a^2}=-12a^2<0$) . therefore $$a-ble 0$$and $$(a-b)^3le 0$$
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$$frac{1}{4}(a-b)(b^2-a^2) + frac{1}{3}(a^3 - b^3){={1over 12}(a-b)Bigg(7b^2+a^2+4abBigg)}$$also $7b^2+a^2+4ab$ is always non-negative (as a function of $b$ and for different values of $a$ where $7b^2+a^2+4ab=0$ iff $a=b=0$) since for $ane 0$ we have $Delta<0$ ($Delta={4^2a^2-4times 7a^2}=-12a^2<0$) . therefore $$a-ble 0$$and $$(a-b)^3le 0$$
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
edited Dec 9 '18 at 18:03
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 9 '18 at 17:56
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
$begingroup$
but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 19:43
|
show 2 more comments
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
$begingroup$
but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 19:43
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
what is the condition known as Δ? why does Δ<0 mean the function is non-negative
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 17:59
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
I edited my answer to remove the ambiguity...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 18:03
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
i am still confused as to how triangle being less than 0 means its non negative..
$endgroup$
– pablo_mathscobar
Dec 9 '18 at 18:17
$begingroup$
If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
$endgroup$
– Mostafa Ayaz
Dec 9 '18 at 19:36
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If a quadratic function has $Delta<0$ it is always positive or negative for example $x^2+2x+2=(x+1)^2+1$ and since $delta=-4<0$ it is either negative or always positive. Since for $x=0$ we have $x^2+2x+2=2$ therefore $x^2+2x+2>0$ for all $xin Bbb R$. You can construct more examples too and check out the delta for them...
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– Mostafa Ayaz
Dec 9 '18 at 19:36
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but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
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– pablo_mathscobar
Dec 9 '18 at 19:43
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but how does that tell us that 7b^2 + a^2 + 4ab is non negative? all it says is that it is either positive or negative as you claim. i cant see how you can get information about a-b from that
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– pablo_mathscobar
Dec 9 '18 at 19:43
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Note there is the obvious factor $a-b$ in the LHS, so ...
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– user10354138
Dec 9 '18 at 17:49