# The sum of two numbers in 38 and the sum of their squares is 730, what are the two numbers?

*print*Print*list*Cite

Let the numbers be x and y.

Given that the sum of the numbers is 38.

Then, we will write:

x + y = 38 ..............(1)

Also, given that the sum of the squares is 730.

Then, we will write:

x^2 + y^2 = 730 ..............(2)

Now we will solve the system.

First we will rewrite (1).

==> x + y = 38 ==> x = 38 - y ...........(3)

Now let us substitute into (2).

==> x^2 + y^2 = 730

==> (38-y)^2 + y^2 = 730

==> 1444 - 76 y +y^2 + y^2 = 730

==> 2y^2 - 76y + 1444 - 730 = 0

==> 2y^2 - 76y + 714 = 0

Now we will divide by 2:

==> y^2 - 38y + 357 = 0

Now we will factor:

==> ( y - 17) ( y- 21) = 0

==> y=1 = 17 ==> x1 = 38-17 = 21

==> y2= 21 ==> x2 = 38 - 21 = 17

**Then, the numbers are 17 and 21.**

The sum of two numbers is given as 38 and the sum of their squares is 730. Let the numbers be denoted by A and B.

So we have A + B = 38 and A^2 + B^2 = 730.

A + B = 38

=> A = 38 – B

Substitute this value of A in A^2 + B^2 = 730

=> (38 – B) ^2 + B^2 = 730

=> 38^2 – 76B + B^2 + B^2 = 730

=> 1444 – 76B + 2B^2 = 730

=> 2B^2 – 76B + 714 = 0

=> B^2 – 38B + 357 = 0

=> B^2 – 21B – 17B + 357 =0

=> B (B – 21) – 17(B – 21) =0

=> (B – 17) (B – 21) =0

So B can take the values 17 and 21. Conversely A can take the values 21 and 17.

**Therefore the required numbers are 17 and 21.**

Since the sum of the two numbers is 38, we assume the numbers to be x and 38-x.

Their square are x^2 and (38-x)^2. = 38^2-2*38x+x^2.

Given that the sum of their squares is 730, we get the equation:

x^2+(x^2-2*38x+38^2) = 730.

2x^2-76x+38^2-730 = 0.

2x^2-38x+714 = 0.

We divide both sides by 2.

x^2-38x+357 = 0.

(x-21)(x-17) = 0.

Therefore x-21 = 0. Or x-17 = 0.

So x = 21. Or x= 17.