please send fast I m in exam
Answers
Answer:
\begin{gathered}The \: value \: of\\ \: a^{2}+\frac{1}{a^{2}}= 18 \end{gathered}Thevalueofa2+a21=18
Step-by-step explanation:
We \: have \: a -\frac{1}{a}=4---(1)Wehavea−a1=4−−−(1)
\begin{gathered}On \: Squaring \: the \: equation \: (1),\\ \: we \: get \end{gathered}OnSquaringtheequation(1),weget
(a-\frac{1}{a})^{2}=4^{2}(a−a1)2=42
\implies a^{2}+\frac{1}{a^{2}}-2\times a\times \frac{1}{a} = 16⟹a2+a21−2×a×a1=16
/* By algebraic identity:
\boxed {(x-y)^{2}=x^{2}+y^{2}-2xy}(x−y)2=x2+y2−2xy */
\implies a^{2}+\frac{1}{a^{2}}-2 = 16⟹a2+a21−2=16
\implies a^{2}+\frac{1}{a^{2}} = 16+2⟹a2+a21=16+2
\implies a^{2}+\frac{1}{a^{2}} = 18⟹a2+a21=18
Therefore,
\begin{gathered}The \: value \: of\\ \: a^{2}+\frac{1}{a^{2}} = 18 \end{gathered}Thevalueofa2+a21=18
•••♪
Answer:
Hope you will understand
