Question

Solve using formula.

y² + ${\displaystyle \frac{1}{3}}$y = 2.

Answer 1

y² + ${\displaystyle \frac{1}{3}}$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.

${\displaystyle y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

${\displaystyle y=\frac{-1\pm \sqrt{73}}{2\times 3}}$

${\displaystyle y=\frac{-1\pm \sqrt{73}}{6}}$

${\displaystyle y=\frac{-1+\sqrt{73}}{6} \text{or} y=\frac{-1-\sqrt{73}}{6}}$

∴ The roots of the given quadratic equation are ${\displaystyle \frac{-1+\sqrt{73}}{6}\text{and}\frac{-1-\sqrt{73}}{6}}$.