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If $(a+\frac{1}{a})^2+(a-\frac{1}{a})^2=8$, find the value of $(a^2+\frac{1}{a^2})$.
Given: $(a+\frac{1}{a})^2+(a-\frac{1}{a})^2=8$.
To do: To find the value of $(a^2+\frac{1}{a^2})$.
Solution:
$(a+\frac{1}{a})^2= a^2+\frac{1}{a^2}+2$
$(a-\frac{1}{a})^2=a^2+\frac{1}{a^2}-2$
$\Rightarrow ( a+\frac{1}{a})^2+( a-\frac{1}{a})^2=2( a^2+\frac{1}{a^2})$
Substituting values,
$\Rightarrow 8=2( a^2+\frac{1}{a^2})$
$\therefore (a^2+\frac{1}{a^2})=\frac{8}{2}=4$
Updated on: 10-Oct-2022
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