Question Number 173504 by AgniMath last updated on 12/Jul/22
Answered by behi834171 last updated on 12/Jul/22
$${A}=\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\Rightarrow{A}^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}=\mathrm{4}\Rightarrow{A}=\mathrm{2}\:\:.\:\:\blacksquare \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jul/22
$${Welcome}\:\boldsymbol{{Abu}}\:\boldsymbol{{Behi}}\:\boldsymbol{{sir}}\:{after}\:{long}\:{time}! \\ $$
Commented by behi834171 last updated on 12/Jul/22
$${hello}\:{mr}\:{rasheed}.{thank}\:{you}\:{so}\:{much}\:{due}\: \\ $$$${to}\:{warm}\:{welcome}.{mis}\:{you}\:{and}\:{all}\:{of} \\ $$$${my}\:{dear}\:{friends}\:{and}\:{masters}. \\ $$$${I}'{m}\:{here}\:{from}\:{now}\:{to}\:{much}\:{more}\:{times} \\ $$$${after}. \\ $$
Answered by MJS_new last updated on 12/Aug/22
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}\:\Leftrightarrow\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0}\:\Leftrightarrow\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Leftrightarrow\:{x}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2} \\ $$$$ \\ $$$${x}+\frac{\mathrm{1}}{{x}}={a}\:\Rightarrow\:\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=\sqrt{{a}+\mathrm{2}} \\ $$