To solve for $\alpha^2 - \beta^2$ without solving the quadratic equation, we use the relationship between the roots and coefficients of the quadratic equation.

Step 1: Given equation

The quadratic equation is:

$4x^2 - 6x + 3 = 0$

From this equation, we identify:

• $a = 4$
• $b = -6$
• $c = 3$

The roots $\alpha$ and $\beta$ satisfy the following relationships:

1. Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{-6}{4} = \frac{3}{2}$
2. Product of roots: $\alpha \beta = \frac{c}{a} = \frac{3}{4}$

Step 2: Using the difference of squares identity

The difference of squares is given by:

$\alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta)$

We already know $\alpha + \beta = \frac{3}{2}$. Now, to find $\alpha - \beta$, we use the discriminant of the quadratic equation:

$\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta}$

Step 3: Calculating $\alpha - \beta$

Substitute $\alpha + \beta = \frac{3}{2}$ and $\alpha \beta = \frac{3}{4}$:

$\alpha - \beta = \sqrt{\left(\frac{3}{2}\right)^2 - 4\left(\frac{3}{4}\right)}$

$\alpha - \beta = \sqrt{\frac{9}{4} - 3} = \sqrt{\frac{9}{4} - \frac{12}{4}} = \sqrt{-\frac{3}{4}}$

This result indicates that $\alpha - \beta$ is imaginary. Hence, $\alpha^2 - \beta^2$ becomes complex. Using the expression:

$\alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta)$

We find:

$\alpha^2 - \beta^2 = \frac{3}{2} \cdot i\sqrt{\frac{3}{4}}$

$\alpha^2 - \beta^2 = \frac{3i\sqrt{3}}{4}$

Final Answer:

$\alpha^2 - \beta^2 = \frac{3i\sqrt{3}}{4}$

Let me know if you want further clarification on these steps!

Questions for deeper understanding:

1. What is the role of the discriminant in determining if roots are real or complex?
2. How does the difference of squares formula apply to complex numbers?
3. Can you deduce $(\alpha + \beta)^2 + (\alpha - \beta)^2$ using the same quadratic relationships?
4. How would the solution differ if $\alpha$ and $\beta$ were real roots?
5. Can you derive $\alpha^n + \beta^n$ for any integer $n$ using this approach?

Tip: Always examine the discriminant ($b^2 - 4ac$) of a quadratic equation before assuming the nature of its roots!