Question

plz answer both the questions.....​

Answer 1

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION:}}}$

Let f(x)= 3x^2 +8x+(2k+1)

$\alpha \: and \: beta \: \: be \: its \: zeroes$

Here, a= 3

b= 8

c= 2k+1

Given,

$\alpha = 7 \beta .......(1) \\ = > sum \: of \: the \: roots \: \alpha + \beta = - \frac{b}{a} \\ \\ = > 7 \beta + \beta = - \frac{8}{3} \: \: (using \: (1)) \\ \\ = > 8 \beta = - \frac{8}{3} \\ \\ = > \beta = - \frac{1}{3}$

Putting Beta= -1/3 in (1), we have

$\alpha = 7 \times - \frac{1}{3} = - \frac{7}{3} \\ \\ so \: the \: zeroes \: are \: \alpha = - \frac{7}{3} \: and \: \beta = - \frac{1}{3}$

Now,

Product of roots,

$- \frac{1}{3} \times - \frac{7}{3} \\ \\ = > \frac{c}{a} = \frac{7}{9} \\ \\ = > \frac{(2k + 1)}{3} = \frac{7}{9} \\ \\ = > 2k + 1 = \frac{7}{3} \\ \\ = > 6k + 3 = 7 \\ = > 6k = 7 - 3 \\ = > 6k = 4 \\ = > k = \frac{4}{6} = \frac{2}{3} \\ \\ = > k = \frac{2}{3}$

So, the value of k = 2/3

hope it helps ☺️