The given quadratic equation is:

$x^2 - 3\sqrt{3}x - 12 = 0$

1. Nature of Roots:

To determine the nature of the roots, we use the discriminant formula:

$\Delta = b^2 - 4ac$

For the equation $ax^2 + bx + c = 0$, we have:

• $a = 1$
• $b = -3\sqrt{3}$
• $c = -12$

Substitute these values into the discriminant formula:

$\Delta = (-3\sqrt{3})^2 - 4(1)(-12)$ $\Delta = 27 + 48 = 75$

Since $\Delta > 0$, the roots are real and distinct.

2. Finding the Roots:

To find the roots, use the quadratic formula:

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$

Substitute the values of $a$, $b$, and $\Delta$:

$x = \frac{-(-3\sqrt{3}) \pm \sqrt{75}}{2(1)}$ $x = \frac{3\sqrt{3} \pm \sqrt{75}}{2}$

Now, simplify $\sqrt{75}$: $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$

So the roots are: $x = \frac{3\sqrt{3} \pm 5\sqrt{3}}{2}$

Thus, the two roots are:

1. $x_1 = \frac{3\sqrt{3} + 5\sqrt{3}}{2} = \frac{8\sqrt{3}}{2} = 4\sqrt{3}$

2. $x_2 = \frac{3\sqrt{3} - 5\sqrt{3}}{2} = \frac{-2\sqrt{3}}{2} = -\sqrt{3}$

Final Answer:

• Nature of roots: Real and distinct.
• Roots: $x_1 = 4\sqrt{3}$ and $x_2 = -\sqrt{3}$.

Do you need more details or have any questions?

Related Questions:

1. How do you determine the discriminant of a quadratic equation?
2. What happens if the discriminant is negative?
3. How is the quadratic formula derived?
4. Can the quadratic formula solve non-real roots?
5. What are the applications of quadratic equations in real-life problems?

Tip:

Always check the discriminant first to understand the nature of the roots before solving the equation.