Questions: Solve application problems using quadratic equations. 2. A positive real number is 4 less than another. If the sum of the squares of the two numbers is 72 , then find the numbers.

Solution Steps

To solve this problem, we can set up a system of equations based on the given conditions. Let \( x \) be the larger number and \( y \) be the smaller number. According to the problem, \( y = x - 4 \). We also know that the sum of the squares of the two numbers is 72, so \( x^2 + y^2 = 72 \). We can substitute \( y \) in the second equation and solve for \( x \), then find \( y \).

Let \( x \) be the larger number and \( y \) be the smaller number. According to the problem, we have: \[ y = x - 4 \] We also know that the sum of the squares of the two numbers is 72: \[ x^2 + y^2 = 72 \]

Substitute \( y = x - 4 \) into the equation \( x^2 + y^2 = 72 \): \[ x^2 + (x - 4)^2 = 72 \]

Simplify the equation: \[ x^2 + (x^2 - 8x + 16) = 72 \] \[ 2x^2 - 8x + 16 = 72 \] \[ 2x^2 - 8x - 56 = 0 \] \[ x^2 - 4x - 28 = 0 \]

Solve the quadratic equation: \[ x = \frac{4 \pm \sqrt{16 + 112}}{2} \] \[ x = \frac{4 \pm \sqrt{128}}{2} \] \[ x = \frac{4 \pm 8\sqrt{2}}{2} \] \[ x = 2 \pm 4\sqrt{2} \]

For \( x = 2 + 4\sqrt{2} \): \[ y = x - 4 = 2 + 4\sqrt{2} - 4 = -2 + 4\sqrt{2} \]

For \( x = 2 - 4\sqrt{2} \): \[ y = x - 4 = 2 - 4\sqrt{2} - 4 = -2 - 4\sqrt{2} \]

Final Answer