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a) Compute \(x+y\) and \(\sqrt{x^2+y^2}\) when  \(x+5\) and \( y=12\)

b) When is

\(​​​​x+y=\sqrt{x^2+y^2}?\)
When is
\(x+y\neq\sqrt{x^2+y^2}?\)

 Aug 29, 2018

Best Answer 

 #1
avatar+972 
+2

a)

 

You just plug in the values, and evaluate the expression. 

 

b) 

 

\(x+y=\sqrt{x^2+y^2}\\ (x+y)^2=x^2+y^2\\ x^2+2xy+y^2=x^2+y^2\\ 2xy=0\)

 

Therefore, either x, y, or both variables have to be 0 for the expression to be equal. 

 

This works in reverse. If \(x+y\ne\sqrt{x^2+y^2}\), then \(2xy\ne0\).

 

I hope this helped,

 

Gavin. 

 Aug 29, 2018
 #1
avatar+972 
+2
Best Answer

a)

 

You just plug in the values, and evaluate the expression. 

 

b) 

 

\(x+y=\sqrt{x^2+y^2}\\ (x+y)^2=x^2+y^2\\ x^2+2xy+y^2=x^2+y^2\\ 2xy=0\)

 

Therefore, either x, y, or both variables have to be 0 for the expression to be equal. 

 

This works in reverse. If \(x+y\ne\sqrt{x^2+y^2}\), then \(2xy\ne0\).

 

I hope this helped,

 

Gavin. 

GYanggg Aug 29, 2018
 #2
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0

Thank you! This helped!

Guest Aug 29, 2018

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