Question

y²+⅓y=2 solve the equation​

Answer 1

given equation is ,

Y² + 1/3y = 2

now change in its general form ax² + bx + c = 0

so, it will become "3y² + y - 6 = 0

Where, a = 3 , b = 1 , c = -6

now we know the quadratic foumula for finding the value of roots.

Y = [ -b \frac{+}{}\sqrt{b^2 - 4ac}

+

b

2

−4ac

/2a

Y = [ -1 \frac{+}{}\sqrt{1 - 4(3)(-6)}

+

1−4(3)(−6)

/2(3)

Y = [-1 \frac{+}{}\sqrt{73}

+

73

/6

hence the root are y =-1+√73)/6

y=-1+√73)/6