Question 69

If $$x + \frac{1}{x} = 2$$, then find the value of $$x^2 + \frac{1}{x^2}$$.

Solution

$$ (x+ \frac{1}{x})^2 = 4$$

Using the identity$$(a+b)^2=a^2+2 \times a \times b+b^2$$

$$(x+ \frac{1}{x})^2 = x^2+2 \times x \times \frac{1}{x}+ \frac{1}{x^2}$$

$$ x^2+2 \times x \times \frac{1}{x}+ \frac{1}{x^2} =4 $$

$$ x^2+ \frac{1}{x^2} +2=4 $$

solving we get

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

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