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Q. If $x^2+\frac{1}{x^2}=47$ then the value of $x+\frac{1}{x}$ is____

Polynomials and Square Roots of Algebraic Expressions

Solution:

$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2 \times x \times \frac{1}{x} $
$=x^2+\frac{1}{x^2}+2$
$=47+2 $
$=49$
Thus,
$(x+\frac{1}{x})^2=49 $
$\Rightarrow x+\frac{1}{x}=\sqrt{49}=7$