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: Calculate sum and product of zeroes
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: Rewrite \(\alpha^2 + \beta^2\) using 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:
\[\boxed{4}\]