To solve the general quadratic equation where $x = \pm \sqrt{3}$, we can start by using the fact that $x = \pm \sqrt{3}$ are the roots of the quadratic equation. Given that the roots of a quadratic equation $ax^2 + bx + c = 0$ are $r_1$ and $r_2$, we can use the factored form of the quadratic equation:

$a(x - r_1)(x - r_2) = 0$

Here, the roots $r_1 = \sqrt{3}$ and $r_2 = -\sqrt{3}$ lead to:

$a(x - \sqrt{3})(x + \sqrt{3}) = 0$

Now, expand the factored form:

$a\left( (x - \sqrt{3})(x + \sqrt{3}) \right) = 0$

Using the difference of squares identity:

$a(x^2 - (\sqrt{3})^2) = 0$

This simplifies to:

$a(x^2 - 3) = 0$

Finally, expanding gives the general quadratic equation:

$ax^2 - 3a = 0$

Comparing this to the standard form of a quadratic equation $ax^2 + bx + c = 0$, we can identify:

• $a = a$
• $b = 0$ (since there is no $x$-term)
• $c = -3a$

Thus, the general form of the quadratic equation where $x = \pm \sqrt{3}$ is:

$ax^2 - 3a = 0$

If you have any questions or would like more details, feel free to ask!

Here are 5 related questions to expand on this:

1. How do you derive the sum and product of the roots from a quadratic equation?
2. What happens when the discriminant of a quadratic equation is zero?
3. How can you determine the nature of the roots of a quadratic equation?
4. How would the equation change if the roots were $x = \pm 2$ instead of $x = \pm \sqrt{3}$?
5. How can you graph the quadratic equation $x^2 - 3 = 0$?

Tip: The discriminant $b^2 - 4ac$ of a quadratic equation determines the type of roots: real and distinct, real and equal, or complex.