Question

If (a-1/a)=√5 find the value of ••a+1/a and a​

Answer 1

Taking a =x

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

$\tt{X-\dfrac{1}{X}}$=√5

${\sf{\green{\underline{\large{To\:find}}}}}$

$\tt{X+\dfrac{1}{X}}$

• and a

${\sf{\pink{\underline{\Large{Explanation}}}}}$

We know that

$\boxed{\boxed{\sf\red{(a-b)^2=a^2+b^2-2ab}}}$

Therefore

$\tt{(X-\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

Solving

$\tt{X-\dfrac{1}{X}}$=√5

Both side squaring

$\tt{(X-\dfrac{1}{X})^2}$=(√5)²

$\implies\tt{(X-\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}=5$

$\implies\tt{5=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

$\implies\tt{5=X^2+\dfrac{1}{X^2}-2}$

$\implies\tt{5+2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{7=X^2+\dfrac{1}{X^2}}$

Now

$\tt{(X+\dfrac{1}{X})^2}$

$\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{(X-\dfrac{1}{X})^2=7+2\frac{1}{X}\times{X}}$

$\implies\tt{(X-\dfrac{1}{X})^2=7+2}$

$\implies\tt{(X-\dfrac{1}{X})^2=9}$

$\implies\tt{(X-\dfrac{1}{X})=3}$

For x

$\tt{X-\dfrac{1}{X}}$=√5

$\tt{(X-\dfrac{1}{X})=3}$

___________________

2x=√5+3

=>x=$\tt\dfrac{\sqrt{5}+3}{2}$

Answer 2

Formula Used

$\bold{\bf{\blue{(x-y)^2=(x+y)^2-4xy}}}$

Answer

Here , x = a & y = 1/a

$\bold{(a-\frac{1}{a} )^2=(a+\frac{1}{a} )^2-4.a.\frac{1}{a} }\\\\\implies \bold{(\sqrt{5} )^2=(a+\frac{1}{a} )^2-4}\\\\\implies \bold{5+4=(a+\frac{1}{a} )^2}\\\\\implies \bold{(a+\frac{1}{a} )^2=9}\\\\\implies \bold{a+\frac{1}{a}=\sqrt{9} }\\\\\implies \bold{\bf{\blue{a+\frac{1}{a}=\pm 3 }}}$

$\bold{(a-\frac{1}{a} )=\sqrt{5}\;and\;(a+\frac{1}{a} )=\pm 3}\\\\\bold{Now\;adding\;these\;two\;equations\;we\;,\;get\;,}\\\\\bold{2a=\sqrt{5}+3\;and\;2a=\sqrt{5}-3}\\\\\implies \bold{\bf{\blue{a=\frac{\sqrt{5}\pm 3}{2} }}}$