Question

Two nu.here differ by 2. The sum of their squares is 244. Find the numbers

Answer 1

Let's denote the two numbers as \( x \) and \( y \). According to the problem, we have the following two equations:

1. \( x - y = 2 \)
2. \( x^2 + y^2 = 244 \)

From the first equation, we can express \( x \) in terms of \( y \):

\[
x = y + 2
\]

Now, we can substitute this expression for \( x \) into the second equation:

\[
(y + 2)^2 + y^2 = 244
\]

Expanding the first term:

\[
(y^2 + 4y + 4) + y^2 = 244
\]

Combining like terms:

\[
2y^2 + 4y + 4 = 244
\]

Subtracting 244 from both sides gives:

\[
2y^2 + 4y + 4 - 244 = 0
\]

Simplifying further:

\[
2y^2 + 4y - 240 = 0
\]

Now, we can divide the entire equation by 2 to simplify:

\[
y^2 + 2y - 120 = 0
\]

Next, we will factor or use the quadratic formula to solve for \( y \). Factoring, we want two numbers that multiply to \( -120 \) and add to \( 2 \). These numbers are \( 12 \) and \( -10 \):

\[
(y + 12)(y - 10) = 0
\]

Setting each factor to zero gives:

\[
y + 12 = 0 \quad \Rightarrow \quad y = -12
\]
\[
y - 10 = 0 \quad \Rightarrow \quad y = 10
\]

Now we can find the corresponding values for \( x \):

1. If \( y = -12 \):
\[
x = y + 2 = -12 + 2 = -10
\]

2. If \( y = 10 \):
\[
x = y + 2 = 10 + 2 = 12
\]

So the pairs of numbers are:

1. \( (-10, -12) \)
2. \( (12, 10) \)

In conclusion, the two numbers are:

\[
\boxed{(12, 10) \text{ or } (-10, -12)}
\]