Question
If $a -\frac{1}{ a }=10$; find $a +\frac{1}{ a }$
✓
Answer
$a-\frac{1}{a}=10 $
$\left(a-\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2}-2(a)\left(\frac{1}{a}\right) $
$\Rightarrow(10)^2 $
$=a^2+\frac{1}{a^2}-2 $
$=a^2+\frac{1}{a^2} $
$=102$
Now $_t\left( a +\frac{1}{ a ^2}\right)$
$=a^2+\frac{1}{a^2}+2(a)\left(\frac{1}{a}\right) $
$=102+2 $
$=104 $
$=a^2-\frac{1}{a^2} $
$=\sqrt{104} $
$= \pm 2 \sqrt{26} .$
$\left(a-\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2}-2(a)\left(\frac{1}{a}\right) $
$\Rightarrow(10)^2 $
$=a^2+\frac{1}{a^2}-2 $
$=a^2+\frac{1}{a^2} $
$=102$
Now $_t\left( a +\frac{1}{ a ^2}\right)$
$=a^2+\frac{1}{a^2}+2(a)\left(\frac{1}{a}\right) $
$=102+2 $
$=104 $
$=a^2-\frac{1}{a^2} $
$=\sqrt{104} $
$= \pm 2 \sqrt{26} .$
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