does square root of (x^2+y^2) equal to x+y?

Question

I don't think so, because if you let x=1, y=1, square root of (x^2+y^2)=square root of 2, which doesn't equal to x+y

Answer 1

 

 ( X +  Y ) ^2 = ( X + Y ) ( X + Y ) = X ( X +Y ) + Y ( X + Y ) = X^2 + 2X Y + Y ^2

 

    X + Y = √( X + Y ) ^2 = √( X^2 + 2XY + Y ^2 ) ≠ √( X^2 + Y^2)

 

    By Pythagoras Theorem:

 

     X^2 + Y ^2 = Z ^2  / Z is the hypotenuse  of a right triangle with X , Y as legs 

 

    Z =  √(X ^ 2 + Y ^2) 

 

       If letting √( X^2 + Y ^2 ) equals   X +y is in violation of Pythagoras theorem with implication that

        length of the hypotenuse equals the Sum of the length of the legs, rather than square of the

         hypotenuse   being equal the Sum of the square of 2 length.

      One Simple counter example defies that .

      let  X = 3  Y = 4

 

 

         √(X^2 + Y ^2) = √ (3^2 + 4 ^2) = √25 =5

 

 

         ( √X ^2 )+( √Y ^2 ) = X + Y = 4 +3 =7

 

        Students should remember that They can multiply and divide Radicals. but can not add or subtract.

         

 

   

Answer 2

Does √(x^2+y^2) = x+y?

 

No.

 

Let's replace x+y with √((x+y)^2)

 

√(x^2+y^2) =? √((x+y)^2) = √((x+y)(x+y))

 

√(x^2+y^2) =? √(x^2+2xy+y^2)

 

See the 2xy term?

 

The only way for the two sides to be equal is if either x or y was zero.

 

So, in general, √(x^2+y^2) ≠ x+y.

Answer 3

So you are given; √[(x²+y²)=4]

 

we'll break them down to various individual components;

 

√x² + √y² = √4

 

the squares cancel out the square roots and you find the square root of 4

 

x + y = 2

 

 

 

So you are correct.