When to apply negative sign when number is squared Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do we reverse inequality sign when dividing by negative number?Square root of a squared number changes sign, which to apply first?When solving inequalities, does $(x-9)$ count as a negative number?When do I use the 'plus-minus' sign when square rooting both sides of an equation? (example in main body).Restrictions on Factorial UsageWhen do I have to take the solution of a square root of a number with negative sign?Why does the negative sign leave when this expression is simplified?“Adding a negative” and other questions about the minus sign.On Changing The Direction Of The Inequality Sign By Dividing By A Negative Number?When do I apply the distributive property?
Did pre-Columbian Americans know the spherical shape of the Earth?
Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?
Can two people see the same photon?
.bashrc alias for a command with fixed second parameter
Problem with display of presentation
Is there any significance to the prison numbers of the Beagle Boys starting with 176-?
Why weren't discrete x86 CPUs ever used in game hardware?
Why are current probes so expensive?
How do you write "wild blueberries flavored"?
Random body shuffle every night—can we still function?
Can haste grant me and my beast master companion extra attacks?
The test team as an enemy of development? And how can this be avoided?
First paper to introduce the "principal-agent problem"
Did any compiler fully use 80-bit floating point?
What is the proper term for etching or digging of wall to hide conduit of cables
When to apply negative sign when number is squared
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
Does the universe have a fixed centre of mass?
Putting class ranking in CV, but against dept guidelines
Why not use the yoke to control yaw, as well as pitch and roll?
Twin's vs. Twins'
latest version of QGIS fails to edit attribute table of GeoJSON file
How to resize main filesystem
NIntegrate on a solution of a matrix ODE
When to apply negative sign when number is squared
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do we reverse inequality sign when dividing by negative number?Square root of a squared number changes sign, which to apply first?When solving inequalities, does $(x-9)$ count as a negative number?When do I use the 'plus-minus' sign when square rooting both sides of an equation? (example in main body).Restrictions on Factorial UsageWhen do I have to take the solution of a square root of a number with negative sign?Why does the negative sign leave when this expression is simplified?“Adding a negative” and other questions about the minus sign.On Changing The Direction Of The Inequality Sign By Dividing By A Negative Number?When do I apply the distributive property?
$begingroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
$endgroup$
add a comment |
$begingroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
$endgroup$
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
39 mins ago
add a comment |
$begingroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
$endgroup$
I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?
algebra-precalculus recreational-mathematics
algebra-precalculus recreational-mathematics
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 55 mins ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
606
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
39 mins ago
add a comment |
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
Though beware Excel and some similar cases, where=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
39 mins ago
2
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
39 mins ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
39 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Does this question arise from the task to complete a square?
Like $x^2+x=(x+1/2)^2-(1/2)^2$?
In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.
For $x^2+bx+c$ this $a$ would be given by $left(fracb2right)^2$. And so $-a=-left(fracb2right)^2$
And because we want to have this zero, the sign can not be influenced by the square.
Else it would not add up to be $0$.
answered 45 mins ago
CornmanCornman
3,82421233
$endgroup$
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196400%2fwhen-to-apply-negative-sign-when-number-is-squared%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
$endgroup$
add a comment |
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
$endgroup$
add a comment |
$begingroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
$endgroup$
When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 52 mins ago
Minus One-TwelfthMinus One-Twelfth
3,603413
3,603413
add a comment |
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As it is already in the previous answers:
$(-x)^2neq-x^2$
To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 23 mins ago
user665960user665960
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 23 mins ago
user665960user665960
113
answered 23 mins ago
user665960user665960
113
113
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Does this question arise from the task to complete a square?
Like $x^2+x=(x+1/2)^2-(1/2)^2$?
In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.
For $x^2+bx+c$ this $a$ would be given by $left(fracb2right)^2$. And so $-a=-left(fracb2right)^2$
And because we want to have this zero, the sign can not be influenced by the square.
Else it would not add up to be $0$.
answered 45 mins ago
CornmanCornman
3,82421233
$endgroup$
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
add a comment |
$begingroup$
Does this question arise from the task to complete a square?
Like $x^2+x=(x+1/2)^2-(1/2)^2$?
In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.
For $x^2+bx+c$ this $a$ would be given by $left(fracb2right)^2$. And so $-a=-left(fracb2right)^2$
And because we want to have this zero, the sign can not be influenced by the square.
Else it would not add up to be $0$.
answered 45 mins ago
CornmanCornman
3,82421233
$endgroup$
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
add a comment |
$begingroup$
Does this question arise from the task to complete a square?
Like $x^2+x=(x+1/2)^2-(1/2)^2$?
In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.
For $x^2+bx+c$ this $a$ would be given by $left(fracb2right)^2$. And so $-a=-left(fracb2right)^2$
And because we want to have this zero, the sign can not be influenced by the square.
Else it would not add up to be $0$.
answered 45 mins ago
CornmanCornman
3,82421233
$endgroup$
Does this question arise from the task to complete a square?
Like $x^2+x=(x+1/2)^2-(1/2)^2$?
In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.
For $x^2+bx+c$ this $a$ would be given by $left(fracb2right)^2$. And so $-a=-left(fracb2right)^2$
And because we want to have this zero, the sign can not be influenced by the square.
Else it would not add up to be $0$.
answered 45 mins ago
CornmanCornman
3,82421233
answered 45 mins ago
CornmanCornman
3,82421233
answered 45 mins ago
CornmanCornman
3,82421233
answered 45 mins ago
CornmanCornman
3,82421233
3,82421233
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
add a comment |
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
$begingroup$
And the downvote is for what?
$endgroup$
– Cornman
32 mins ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196400%2fwhen-to-apply-negative-sign-when-number-is-squared%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
51 mins ago
$begingroup$
Though beware Excel and some similar cases, where
=-1^2gives1but=0-1^2gives-1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first-as a unary operation taking precedence over exponentiation and the second-as a binary operation with exponentiation taking precedence over it$endgroup$
– Henry
39 mins ago