Question

Solve the following quadratic equation by the formula method:

`y^2 + 1/3y = 2`

Answer 1

`y^2 + 1/3y` = 2

∴ 3y² + y = 6     ......[Multiplying both sides by 3]

∴ 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 - 4"ac"))/(2"a")`

= `(-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`.

Answer 2

`y^2 + 1/3y = 2`

∴ `y^2 + 1/3y - 2 = 0`

Comparing with the standard form ax² + bx + c = 0,

a = 1, b = `1/3`, c = −2.

`b^2 - 4ac = (1/3)^2 - 4(1)(-2)`

= `1/9 + 8`

= `(1 + 72)/9`

= `73/9`

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

= `(-(1/3) +- sqrt(73/9))/(2 xx 1)`

= `(-1/3 +- 1/3 sqrt73)/(2)`

= `(-1 +- sqrt73)/6`

∴ `y = (-1 + sqrt73)/6` or `y = (-1 - sqrt73)/6`

∴ `(-1 + sqrt73)/6`, `(-1 - sqrt73)/6` are the roots of the given quadratic equation.