$\frac{1}{x + 2} = \frac{1}{x^{2}}$

Open in App

Solution

Given: $\frac{1}{x + 2} = \frac{1}{x^{2}}$
First, we rewrite this equation in the standard form of a quadratic equation:
x² = x + 2
=> x² – x – 2 = 0
On splitting the middle term –x as x – 2x, we get:
x² + x – 2x – 2 = 0
=> x(x + 1) – 2(x + 1) = 0
=> (x + 1) (x – 2) = 0
We know that if the product of two numbers is zero then at least one of them must be zero.
Thus, x + 1 = 0 or x – 2 = 0
=> x = –1 or x = 2
Therefore, the solution of the given equation is x = –1,2.

Suggest Corrections

Similar questions

\lim_{x \to 1} \frac{x - 1}{\sqrt{x^{2} + 3 - 2}}

\int \frac{d x}{(1 + x^{2}) \sqrt{p^{2} + q^{2} (t a n^{- 1} x)^{2}}} =

x = 3 - 2 \sqrt{2}

\frac{1}{x}

x + \frac{1}{x}

x - \frac{1}{x}

\sqrt{x} + \frac{1}{\sqrt{x}}

\sqrt{x} - \frac{1}{\sqrt{x}}

x^{2} + \frac{1}{x^{2}}

x^{2} - \frac{1}{x^{2}}

\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = K t a n^{- 1} x + L t a n^{- 1} \frac{x}{2} + C

30 (x^{2} + \frac{1}{x^{2}}) - 77 (x - \frac{1}{x}) - 12 = 0

Join BYJU'S Learning Program