Sum of two numbers is 17, and sum of their squares is 185, what is the number

Suppose that the sum of two numbers is 17, and the sum of their squares is 185. Please help me find the numbers and thank you!

x + y = 17
x² + y² = 185

y = 17 - x

x² + (17 - x)² = 185
x² + 289 - 34x + x² = 185
2x² - 34x + 104 = 0
2(x² - 17x + 52) = 0
2 (x - 13) (x - 4) = 0
x = 13 or 4

Solve for a respective y.

13 + y = 17
or
4 + y = 17

y = 4 or 13.

IN any case, your two numbers are 4 and 13.

Sum of two numbers is 17:

x + y = 17


Sum of their squares is 185

x^2 + y^2 = 185

We got the equation system:

x + y = 17
x^2 + y^2 = 185

From first equation: x = 17 - y, put this to second.

(17 - y)^2 + y^2 = 185

289 - 34y + y^2 + y^2 = 185

After solving this quadratic equation you get y = 4 and y = 13.

Now put these to first equation:

x = 17 - y

1) x = 17 - 13 = 4
2) x = 17 - 4 = 13

So we got two pairs: (x = 17, y=4) and (x = 4, y = 17)

Let one of the number be a
then other number is (17 – a)
a² + (17 – a)² = 185
a² + a² – 34a + 289 = 185
2a² – 34a + 104 = 0
a² – 17a + 52 = 0
(a – 13)(a – 4) = 0
a = 13 or a = 4
other number is 4 or 13
Numbers are 4 and 13
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let n = first number 17-n = second number

n^2 + (17-n)^2 = 185

n^2 +289 - 34n +n^2 = 185

2n^2 -34n +104 = 0

n^2 -17n +52 =0

(n-13)(n-4) = 0

The two numbers are 4 and 13

x + y = 17

x^2 + y^2 = 185
(x + y)^2 = 185 + 2xy
289 = 185 + 2xy
xy = 52

y = 17 - x

x(17 - x) = 52
17x - x^2 = 52
x^2 - 17x + 52 = 0
(x - 13)(x - 4)
x = 13, x = 4
y = 4, y = 13

therefore, numbers are 4 and 17

x + y = 17 → y = 17 - x

... x^2 + y^2 = 185
or x^2 + (17-x)^2 = 185
or x^2 + x^2 - 34x + 289 = 185
or 2x^2 - 34x + 104 = 0
or x^2 - 17x + 52 = 0
or (x - 4)(x - 13) = 0
or { x,y } = { 4,13 }

a+b=17
a^2+b^2=185
the numbers are 13 and 4