The given quadratic equation is:

$ax^2 + 4ax + c = 0$

To find the roots of this equation, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the given equation:

• $a$ (coefficient of $x^2$) = $a$
• $b$ (coefficient of $x$) = $4a$
• $c$ (constant term) = $c$

Substituting these values:

$x = \frac{-4a \pm \sqrt{(4a)^2 - 4(a)(c)}}{2(a)}$

$x = \frac{-4a \pm \sqrt{16a^2 - 4ac}}{2a}$

$x = \frac{-4a \pm 2\sqrt{4a^2 - ac}}{2a}$

$x = \frac{-4a}{2a} \pm \frac{2\sqrt{4a^2 - ac}}{2a}$

$x = -2 \pm \frac{\sqrt{4a^2 - ac}}{a}$

Thus, the roots of the equation are:

$x = -2 \pm \frac{\sqrt{4a^2 - ac}}{a}$

Would you like a step-by-step breakdown of how I solved this?

Related Questions:

1. What conditions on $a$ and $c$ ensure real roots?
2. How does the discriminant affect the nature of the roots?
3. How do we factorize this quadratic if possible?
4. What happens when $c = 4a^2$?
5. Can this equation have equal roots? If so, when?

Tip: The discriminant $\Delta = 16a^2 - 4ac$ determines the nature of the roots. If $\Delta > 0$, the roots are real and distinct; if $\Delta = 0$, they are real and equal; if $\Delta < 0$, they are complex.