Identify the given roots and the general form of the quadratic equation.

Right, let's go. Step 1: Identify the given roots and the general form of the quadratic equation. The given quadratic equation is $x^2 + bx + c = 0$. The roots are $x_1 = \frac{-7+\sqrt{61}}{2}$ and $x_2 = \frac{-7-\sqrt{61}}{2}$. For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $x_1 + x_2 = -\frac{b}{a}$ and the product of the roots is $x_1 \cdot x_2 = \frac{c}{a}$. In this case, $a=1$. Step 2: Calculate the sum of the roots to find the value of $b$. $$x_1 + x_2 = \frac{-7+\sqrt{61}}{2} + \frac{-7-\sqrt{61}}{2}$$ $$x_1 + x_2 = \frac{-7+\sqrt{61} - 7 - \sqrt{61}}{2}$$ $$x_1 + x_2 = \frac{-14}{2}$$ $$x_1 + x_2 = -7$$ Since $x_1 + x_2 = -\frac{b}{a}$ and $a=1$: $$-7 = -b$$ $$b = 7$$ Step 3: Calculate the product of the roots to find the value of $c$. $$x_1 \cdot x_2 = \left(\frac{-7+\sqrt{61}}{2}\right) \cdot \left(\frac{-7-\sqrt{61}}{2}\right)$$ This is in the form $(A+B)(A-B) = A^2 - B^2$, where $A = -7$ and $B = \sqrt{61}$. $$x_1 \cdot x_2 = \frac{(-7)^2 - (\sqrt{61})^2}{2 \cdot 2}$$ $$x_1 \cdot x_2 = \frac{49 - 61}{4}$$ $$x_1 \cdot x_2 = \frac{-12}{4}$$ $$x_1 \cdot x_2 = -3$$ Since $x_1 \cdot x_2 = \frac{c}{a}$ and $a=1$: $$-3 = c$$ The values are $b=7$ and $c=-3$. The final answer is $\boxed{b=7, c=-3}$. Send me the next one 📸