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Solving a Quadratic Equation by Factorization Method

Factorize:𝑥2...

Question

Factorize:
$(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$

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Solution

Given: $(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$

$(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$
$= (x^{2} + \frac{1}{x^{2}} + 2) - 2 - 4 (x + \frac{1}{x}) + 6$
$= (x + \frac{1}{x})^{2} - 4 (x + \frac{1}{x}) + 4$
$= (x + \frac{1}{x} - 2)^{2}$ $[∵ a^{2} + b^{2} - 2 a b = (a - b)^{2}]$

Hence, $(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$ $= (x + \frac{1}{x} - 2)^{2}$

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Solving a Quadratic Equation by Factorization Method

Standard X Mathematics

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