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