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प्रश्न
Solve the following quadratic equations by formula method.
`y^2 + 1/3y` = 2
उत्तर
`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 y = `(-1 - sqrt(73))/6`.
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