Question

If the roots of 5n^2+7kn+3+0 are reciprocal of the roots of the equation 3n^2+(8-k)n+5 then the value of k is

Answer 1

Step-by-step explanation:

Text Solution

0

1

2

5

Answer :

D

Solution :

Let the roots be

Then, <br> product of roots

I hope it helps. Have a great day.

Answer 2

Step-by-step explanation:

Let the roots of $5n^2+7kn+3=0\:are\:\alpha\:and\:\beta\\\\\alpha+\beta=\frac{-7k}{5}...eq1\\\\\alpha\times\beta=\frac{3}{5}...eq2$

Now ATQ,roots of eq2 are

$3n^2+(8-k)n+5=0\:are\:\frac{1}{\alpha}\:and\:\frac{1}{\beta}\\\\\frac{1}{\alpha}+\frac{1}{\beta}=\frac{k-8}{3}...eq3\\\\\frac{1}{\alpha\times\beta}=\frac{5}{3}...eq4$

From eq3,take LCM

$\frac{\alpha+\beta}{\alpha\beta}=\frac{k-8}{3}$

Put value of eq1 and eq2 in eq 5

$\frac{-7k\times5}{5\times3}=\frac{k-8}{3}\\\\\frac{-7k}{3}=\frac{k-8}{3}\\\\-7k=k-8\\\\-7k-k=-8\\\\-8k=-8\\\\k=1$

k=1,for the given condition of roots.

Hope it helps you.