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प्रश्न
Solve the following quadratic equation.
`1/(x + 5) = 1/x^2`
योग
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उत्तर
`1/(x + 5) = 1/x^2`
∴ x2 = x + 5
∴ x2 − x − 5 = 0
Comparing the above equation with ax2 + bx + c = 0, we get,
a = 1, b = −1, c = −5
∴ b2 − 4ac = (−1)2 − 4 × 1 × −5 = 1 + 20 = 21
`x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`x = (-(-1) +- sqrt(21))/(2 × 1)`
`x = (1 +- sqrt(21))/(2)`
`x = (1 + sqrt(21))/(2) or x = (1 - sqrt(21))/2`
∴ The roots of the given quadratic equation are `(1 + sqrt(21))/(2) "and" (1 - sqrt(21))/2`.
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