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?
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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 1 hour 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 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
$endgroup$
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour 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
47 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 1 hour 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 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
asked 1 hour ago
JohnJohnyPapaJohnJohnJohnyPapaJohn
606
606
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour 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
47 mins ago
add a comment |
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour 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
47 mins ago
2
2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour ago
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour 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
47 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
47 mins ago
add a comment |
2 Answers
2
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 1 hour 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 32 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 |
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2 Answers
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2 Answers
2
active
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$begingroup$
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
3,603413
answered 1 hour 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 32 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 32 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 32 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 32 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 32 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 32 mins ago
user665960user665960
113
answered 32 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.
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2
$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
1 hour 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
47 mins ago