If the sum of squares of zeroes of the polynomial x² + 9x + 3k is 21, then find the value of k.

The polynomial is f(x) = $$x^2 + 9x + 3k$$​, and the sum of squares of its zeroes is 21.

For a quadratic polynomial of the form $$ax^2 + bx + c:$$​​

If α and β are the zeroes, then:

Sum of the zeroes, $$\alpha + \beta = -\frac{b}{a}$$​​

Product of the zeroes, $$\alpha \beta = \frac{c}{a}$$​​

The sum of the squares of the zeroes is given by: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$$​​

Calculating α+β and αβ:

​$$\alpha + \beta = -\frac{b}{a} = -\frac{9}{1} = -9$$​​

​$$\alpha \beta = \frac{c}{a} = \frac{3k}{1}$$​​

Substituting into the formula for $$\alpha^2 + \beta^2:$$​​

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

​$$\alpha^2 + \beta^2 = (-9)^2 - 2 \times 3k$$​​

​$$\alpha^2 + \beta^2 = 81 - 6k$$​​

Given that $$\alpha^2 + \beta^2 = 21:$$​​

81 - 6k = 21

6k = 81 – 21

6k = 60

k = $$\frac{60}{6}$$​ = 10