1. **State the problem:** Find integer values for $a$ and $c$ such that the quadratic equation $ax^2 - 5x + c = 0$ has two imaginary solutions. 2. **Recall the quadratic formula and discriminant:** The solutions of $ax^2 + bx + c = 0$ are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots: - If $\Delta > 0$, two distinct real roots. - If $\Delta = 0$, one real root (repeated). - If $\Delta < 0$, two imaginary (complex conjugate) roots. 3. **Apply to our equation:** Here, $a = a$, $b = -5$, and $c = c$. The discriminant is $$\Delta = (-5)^2 - 4 \cdot a \cdot c = 25 - 4ac$$ 4. **Condition for imaginary roots:** We want $\Delta < 0$, so $$25 - 4ac < 0 \implies 4ac > 25 \implies ac > \frac{25}{4} = 6.25$$ 5. **Choose integer values for $a$ and $c$ satisfying $ac > 6.25$:** - For example, $a = 1$, $c = 7$ since $1 \times 7 = 7 > 6.25$ - Or $a = 2$, $c = 4$ since $2 \times 4 = 8 > 6.25$ 6. **Write the quadratic equation:** Using $a=1$, $c=7$, the equation is $$1x^2 - 5x + 7 = 0$$ 7. **Verify discriminant:** $$\Delta = 25 - 4 \times 1 \times 7 = 25 - 28 = -3 < 0$$ This confirms two imaginary solutions. **Final answer:** A possible quadratic equation is $$x^2 - 5x + 7 = 0$$ with $a=1$ and $c=7$ producing two imaginary solutions.