Question

find roots of y-1/3y=1/6​

Answer 1

Solution:-

$\rm \to \: y - \dfrac{1}{3y} = \dfrac{1}{6}$

Now simplify

$\rm \to \: \dfrac{3y(y) - 1}{3y} = \dfrac{1}{6}$

$\rm \to \: \dfrac{3 {y}^{2} - 1 }{3y} = \dfrac{1}{6}$

Using cross multiplication

$\rm \to \: 6(3 {y}^{2} - 1) = 3y(1)$

$\rm \to \: 18 {y}^{2} - 6 = 3y$

$\rm \to \: 18 {y}^{2} - 3y - 6 = 0$

$\rm \to \: 3(6 {y}^{2} - y - 2) = 0$

$\rm \to \: 6 {y}^{2} - y - 2 = 0$

Now factories the equation

$\rm \to \: 6 {y}^{2} - y - 2 = 0$

$\rm \to \: 6 {y}^{2} -4y + 3y - 2 = 0$

$\rm \to \: 2y(3y - 2) + 1(3y - 2) = 0$

$\rm \to \: (2y + 1)(3y - 2) = 0$

$\rm \to \: 2y + 1 = 0 \: \: and \: \: 3y - 2 = 0$

$\rm \to \: y = \dfrac{ - 1}{2} \: and \: y = \dfrac{2}{3}$

So roots of given equation is

$\rm \to \: y = \dfrac{ - 1}{2} \: and \: y = \dfrac{2}{3}$

Answer 2

Answer:

$y - \frac{1}{3y} = \frac{1}{6}$

take L.C.M

$\to \frac{3 {y}^{2} - 1 }{3y} = \frac{1}{6}$

$\to3 {y}^{2} - 1 = \frac{3y}{6} = \frac{y}{2}$

cross multiply

$\to6 {y}^{2} - 2 = y$

$\to6 {y}^{2} - y - 2 = 0$

solving the quadratic equation we get

$\to2y(3y - 2) + 1(3y - 2) = 0$

$\to(2y + 1)(3y - 2) = 0 \\ \to \: y = \frac{ - 1}{2} , \frac{2}{3}$