Question

solve this question​

Answer 1

The value of \bold{a+\frac{1}{a}}a+a1 is 5.

Step-by-step explanation:

Given a^{2}+\frac{1}{a^{2}}=23a2+a21=23

We know that by formula, \bold{(a+b)^{2}=a^{2}+b^{2}+2 \times a \times b}(a+b)2=a2+b2+2×a×b

Bring a^{2}+\frac{1}{a^{2}}a2+a21 in the above format, hence the equation becomes,

\begin{gathered}\begin{aligned}\left(a+\frac{1}{a}\right)^{2} &=a^{2}+\frac{1}{a^{2}}+2 \times a^{2} \times \frac{1}{a^{2}} \\\left(a+\frac{1}{a}\right)^{2} &=23+2 \end{aligned}\end{gathered}(a+a1)2(a+a1)2=a2+a21+2×a2×a21=23+2

Hence \bold{\left(a+\frac{1}{a}\right)^{2}=25}(a+a1)2=25

Now by taking square root on both sides, the equation becomes,

\bold{a+\frac{1}{a}=5}a+a1=5

Step-by-step explanation:

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