Question

if a^2 + 1/ a^2 =23, find the value of (a + 1/a)​

Answer 1

$\rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) }^{2} = {a}^{2} + {( \dfrac{1}{a} )}^{2} +( 2 \times a \times \dfrac{1}{a}) \\ \\ \\ \rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) }^{2} = \rm {a}^{2} + {( \dfrac{1}{a} )}^{2} +2 \\ \\ \rm it \: is \: given \: that \: {a}^{2} + { \dfrac{1}{ {a}^{2} } } = 23 \\ \\\\ \rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) }^{2} = \rm 23+2 \\\\ \rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) }^{2} = \rm 25\\\\ \rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) } = \rm \sqrt{25} \\\\ \rm \rightarrow { \bigg( {a}+ \dfrac{1}{a} \bigg) } = \rm \pm 5$

Answer : ±5

Learn more :

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$