Question

Solve using formula.

y² + `1/3`y = 2.

Answer 1

y² + `1/3`y = 2

Multiplying both sides by 3,

∴ 3y² + y = 6

∴ 3y² + y − 6 = 0

Comparing the above equation with ay² + by + c = 0, we get,

a = 3, b = 1, c = −6

∴ b² – 4ac = (1)² – 4 × 3 × (−6) = 1 + 72 = 73.

`y = (-b +- sqrt(b^2 - 4ac))/(2a)`

`y = (-1 +- sqrt(73))/(2 × 3)`

`y = (-1 +- sqrt(73))/(6)`

`y = (-1 + sqrt(73))/(6)     "or"      y = (-1 - sqrt(73))/(6)`

∴ The roots of the given quadratic equation are `(-1 + sqrt(73))/(6) "and" (-1 - sqrt(73))/(6)`.