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Solve the following quadratic equation. 1x+5=1x2 - Algebra

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Question

Solve the following quadratic equation.

`1/(x + 5) = 1/x^2`

Sum

Solution

`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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Chapter 2: Quadratic Equations - Problem Set 2 [Page 54]

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Balbharati Algebra (Mathematics 1) [English] 10 Standard SSC Maharashtra State Board
Chapter 2 Quadratic Equations
Problem Set 2 | Q 7.1 | Page 54

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