Question

quadratic equation -
1/x+5 = 1/x^2
how do you solve this?

Answer 1

Hey there!

Simplify the equation to solve it by quadratic formula:

$\bf{\drfac{1}{x + 5} = \dfrac{1}{x^2}}$

$\bf{x^2 = x + 5}$

$\bf{x^2 - 5 = x + 5 - 5}$

$\bf{x^2 - 5 = x}$

$\bf{x^2 - 5 - x = x - x}$

$\bf{\underline{x^2 - x - 5 = 0}}$

Solving it by applying quadratic formula for equation "x^2 - x - 5 = 0"

For a required quadratic equation of $\bf{ax^2 + bx + c = 0}$ the solutions for which can be represented by the quadratic formula of :

$\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = 1,  b = - 1,  c = - 5.

Solving for positive and negative values respectively:

$\bf{x_{1} = \dfrac{- (- 1) + \sqrt{(- 1)^2 - 4(1)(- 5)}}{2(1)}}$

$\bf{x_{1} = \dfrac{1 + \sqrt{21}}{2 \times 1}}$

$\bf{x_{1} = \dfrac{1 + \sqrt{21}}{2}}$

$\bf{\therefore \quad x_{1} = \dfrac{1 + \sqrt{21}}{2}}$

For the second solution in negative form of equation:

$\bf{x_{2} = \dfrac{- (- 1) - \sqrt{(- 1)^2 - 4(1)(- 5)}}{2(1)}}$

$\bf{x_{2} = \dfrac{1 - \sqrt{21}}{2 \times 1}}$

$\bf{x_{2} = \dfrac{1 - \sqrt{21}}{2}}$

$\bf{\therefore \quad x_{2} = \dfrac{1 - \sqrt{21}}{2}}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_{1} = \dfrac{1 + \sqrt{21}}{2}}}}$

$\boxed{\underline{\bf{x_{2} = \dfrac{1 - \sqrt{21}}{2}}}}$

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!