Numbers Finder by Difference and Sum of Squares

Input the difference and sum of squares, and instantly find the two numbers.

How to Calculate Two Numbers Using Their Difference and Sum of Squares

Given:

Steps:

Set up relationships: \( x - y = D, \quad x^2 + y^2 = P \)
Express one number in terms of the other: \( x = y + D \)
Substitute into the sum of squares formula: \( (y + D)^2 + y^2 = P \)
Expand and simplify: \( y^2 + 2Dy + D^2 + y^2 = P \)
Combine like terms and rewrite as a quadratic equation: \( 2y^2 + 2Dy + D^2 - P = 0 \) \( y^2 + Dy + \frac{D^2 - P}{2} = 0 \)
Solve for \( y \) using the quadratic formula: \( y = \frac{-D \pm \sqrt{D^2 - 2(D^2 - P)}}{2} \)
Find \( x \) using \( x = y + D \).

Solution:

1. Solve the quadratic equation:

\( y^2 + 2y + \frac{2^2 - 52}{2} = 0 \)

\( y^2 + 2y - 24 = 0 \)

2. Calculate the discriminant:

\( \sqrt{2^2 - 4 \cdot 1 \cdot (-24)} = \sqrt{4 + 96} = \sqrt{100} = 10 \).

3. Solve for \( y \):

\( y = \frac{-2 \pm 10}{2} = 4 \text{ or } -6 \)

4. Compute \( x \):

\( x = y + 2 \)

\( x = 6 \) or \( -4 \)

Result: The numbers are (6, 4) or (-4, -6).

Solution:

1. Solve the quadratic equation:

\( y^2 + 11y + \frac{11^2 - 145}{2} = 0 \)，即 \( y^2 + 11y - 12 = 0 \).

2. Calculate the discriminant:

\( \sqrt{11^2 - 4 \cdot 1 \cdot (-12)} = \sqrt{121 + 48} = \sqrt{169} = 13 \).

3. Solve for \( y \):

\( y = \frac{-11 \pm 13}{2} = 1 \text{ or } -12 \)

4. Compute \( x \):

\( x = y + 11 \)

\( x = 12 \) or \( -1 \)

Result: The numbers are (12, 1) or (-1, -12).