Question
Solve the following quadratic equation.
$\frac{1}{x+5}=\frac{1}{x^2}$

Answer


$\begin{array}{l}x^2=x+5 \\ \Rightarrow x^2-x-5=0 \\ \Rightarrow x^2-x-5=0 \text { compare with } a x^2+b x+c=0 \\ \Rightarrow a=1, b=-1 \text { and } c=-5 \\ \therefore b^2-4 a c=-1^2-4(1)(-5) \\ =1+20 \\ =21\end{array}$
$\begin{array}{l}x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ \Rightarrow x=\frac{1 \pm \sqrt{21}}{2 \times 1} \\ \Rightarrow x=\frac{1 \pm \sqrt{21}}{2} \\ \Rightarrow x=\frac{1+\sqrt{21}}{2} \text { or } x=\frac{1+\sqrt{21}}{2}\end{array}$

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