Question

\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

Hint:

Use identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) and convert all to same denominator.

Answer 1

Given:
Polynomial: \(3x^2 - 8x + k = 0\)
Zeroes: \(\alpha, \beta\)
Condition: \(\alpha^2 + \beta^2 = \frac{40}{9}\)

Step 1: Use sum and product of zeroes formulas
Sum of zeroes:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-8}{3} = \frac{8}{3} \]
Product of zeroes:
\[ \alpha \beta = \frac{c}{a} = \frac{k}{3} \]

Step 2: Express \(\alpha^2 + \beta^2\) in terms of sum and product
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substitute values:
\[ \frac{40}{9} = \left(\frac{8}{3}\right)^2 - 2 \times \frac{k}{3} \] \[ \frac{40}{9} = \frac{64}{9} - \frac{2k}{3} \]

Step 3: Solve for \(k\)
\[ \frac{2k}{3} = \frac{64}{9} - \frac{40}{9} = \frac{24}{9} = \frac{8}{3} \] Multiply both sides by \(\frac{3}{2}\):
\[ k = \frac{8}{3} \times \frac{3}{2} = 4 \]

Final Answer:
\[ \boxed{4} \]