Question

y²+1/3y=2 Solve using formula.

Answer 1

y² + 1/3 y = 2
first of all resolve it in general form ay² + by + c = 0 .
e.g., y² + 1/3 y = 2
=> 3y² + y = 2 × 3
=> 3y² + y - 6 = 0

if ay² + by + c = 0
then, y = {-b ± √(b² - 4ac)}/2a use it here,

3y² + y - 6 = 0
so, y = {-1 ± √(1² - 4.3(-6))}/2(3)
= {-1 ± √(1 + 72)}/6
= (-1 ± √73)/6

hence, roots are (-1 + √73)/6 and (-1 - √73)/6

Answer 2

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 $\frac{+}{}\sqrt{b^2 - 4ac}$]/2a

Y = [ -1 $\frac{+}{}\sqrt{1 - 4(3)(-6)}$]/2(3)

Y = [-1 $\frac{+}{}\sqrt{73}$]/6

hence, the roots are, Y = (-1 + $\sqrt{73}$)/6

Y = (1 - $\sqrt{73}$)/6