Certainly! Let's delve into the problem of finding the value of \( \alpha^2 + \beta^2 \) where \( \alpha \) and \( \beta \) are the zeros of the quadratic polynomial \( 3x^2 - 2x - A \).
Understanding the Quadratic Polynomial
The standard form of a quadratic polynomial is given by:
\[ ax^2 + bx + c = 0 \]
For our polynomial:
- \( a = 3 \)
- \( b = -2 \)
- \( c = -A \)
Using Vieta's Formulas
According to Vieta's formulas, the sum and product of the roots (zeros) \( \alpha \) and \( \beta \) can be expressed as:
- \( \alpha + \beta = -\frac{b}{a} = -\frac{-2}{3} = \frac{2}{3} \)
- \( \alpha \beta = \frac{c}{a} = \frac{-A}{3} \)
Finding \( \alpha^2 + \beta^2 \)
To find \( \alpha^2 + \beta^2 \), we can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
- Calculate \( (\alpha + \beta)^2 \):
\[ \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
- Now calculate \( 2\alpha\beta \):
\[ 2\alpha\beta = 2 \left(\frac{-A}{3}\right) = \frac{-2A}{3} \]
Putting it all together:
\[ \alpha^2 + \beta^2 = \frac{4}{9} - \left(\frac{-2A}{3}\right) \]
To combine these fractions, convert \( \frac{-2A}{3} \) to have a common denominator:
\[ \frac{-2A}{3} = \frac{-6A}{9} \]
Thus:
\[ \alpha^2 + \beta^2 = \frac{4}{9} + \frac{6A}{9} = \frac{4 + 6A}{9} \]
Final Result
The value of \( \alpha^2 + \beta^2 \) is:
\[ \alpha^2 + \beta^2 = \frac{4 + 6A}{9} \]