Sum Product 8952Bd

1. **State the problem:** Find two numbers whose sum is $\frac{1}{4}$ and whose product is $-\frac{1}{8}$. 2. **Use the quadratic equation approach:** Let the two numbers be $x$ and $y$. We know: $$x + y = \frac{1}{4}$$ $$xy = -\frac{1}{8}$$ 3. **Form a quadratic equation:** If $x$ and $y$ are roots of the quadratic equation, then the equation is: $$t^2 - (x+y)t + xy = 0$$ Substitute the known sums and products: $$t^2 - \frac{1}{4}t - \frac{1}{8} = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-\frac{1}{4}$, and $c=-\frac{1}{8}$. Calculate the discriminant: $$\Delta = \left(-\frac{1}{4}\right)^2 - 4 \times 1 \times \left(-\frac{1}{8}\right) = \frac{1}{16} + \frac{1}{2} = \frac{1}{16} + \frac{8}{16} = \frac{9}{16}$$ 5. **Find the roots:** $$t = \frac{-\left(-\frac{1}{4}\right) \pm \sqrt{\frac{9}{16}}}{2 \times 1} = \frac{\frac{1}{4} \pm \frac{3}{4}}{2}$$ Calculate each root: - For the plus sign: $$t = \frac{\frac{1}{4} + \frac{3}{4}}{2} = \frac{1}{2}$$ - For the minus sign: $$t = \frac{\frac{1}{4} - \frac{3}{4}}{2} = \frac{-\frac{1}{2}}{2} = -\frac{1}{4}$$ 6. **Answer:** The two numbers are $\frac{1}{2}$ and $-\frac{1}{4}$.