Question

solve the following quadratic equations by factorization method

$1 \div x + 2 = 1 \div x { }^{2}$
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Answer 1

Step-by-step explanation:

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so here w go

so 1/x+2=1/x square

ie x square=x+2

ie x square-x-2=0

so x square-2x+x-2=0

ie x(x-2)+1 (x-2)=0

(x-2)(x+1)=0

ie x-2=0 or x+1=0

ie x=2 or x=-1

hence 2 and -1 are required roots of above quadratic equation

Answer 2

Step-by-step explanation:

Question:-

Solve it

$\\ \tt{:}\twoheadrightarrow \dfrac{1}{x+2}=\dfrac{1}{x^2}$

SOLUTION:-

$\\ \tt{:}\twoheadrightarrow \frac{1}{x + 2} = \frac{1}{ {x}^{2} } \\ \bf \bigstar\: using \: cross \: multiplication\\ \\ \tt{:}\twoheadrightarrow {x}^{2} = x + 2 \\ \\ \tt{:}\twoheadrightarrow {x}^{2} - x - 2 = 0 \\ \\ \tt{:}\twoheadrightarrow {x}^{2} - 2x + x - 2 = 0 \\ \\ \tt{:}\twoheadrightarrow x(x - 2) + 1(x - 2) = 0 \\ \\ \tt{:}\twoheadrightarrow (x - 2)(x + 1) = 0 \\ \\ \tt{:}\twoheadrightarrow (x - 2) = 0 \:or \: (x + 1) = 0 \\ \\ \tt{:}\twoheadrightarrow x = 2 \: or \: x = - 1$