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प्रश्न
Solve using formula.
y2 + `1/3`y = 2.
उत्तर
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)`.
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