Question

if a=4/3-√5,then the value of (a+4/a) is​

Answer 1

$\large\bold{\underline{\underline{Given:-}}}$

$a = \frac{4}{3 - \sqrt{5} }$

$\large\bold{\underline{\underline{To\:find:-}}}$

$(a + \frac{4}{a} )...(i)$

$\large\bold{\underline{\underline{Solution:-}}}$

Let us first solve for $a$.

:⟼  $\frac{4}{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} }$

:⟼  $\frac{4(3 + \sqrt{5} )}{ ({3})^{2} - ({ \sqrt{5} })^{2} }$

:⟼  $\frac{12 +4 \sqrt{5} }{9 - 5}$

:⟼  $\frac{4(3 + \sqrt{5} )}{4}$

:⟼  $3 + \sqrt{5} ...(ii)$

Now that we have the value of $a$, let us solve for $\frac{4}{a}$.

:⟼  $\frac{4}{3 + \sqrt{5} }$

:⟼  $\frac{4}{3 + \sqrt{5} } \times \frac{3 - \sqrt{5} }{3 - \sqrt{5} }$

:⟼  $\frac{4(3 - \sqrt{5} )}{ ({3})^{2} - ({ \sqrt{5} })^{2} }$

:⟼  $\frac{12 - 4 \sqrt{5} }{9 - 5}$

:⟼  $\frac{4(3 - \sqrt{5}) }{4}$

:⟼  $3 - \sqrt{5} ...(iii)$

Substituting $eqn.(ii)$ and $eqn.(iii)$ in $eqn.(i)$, we have

:⟼  $a + \frac{4}{a}$

:⟼  $3 + \sqrt{5} + 3 - \sqrt{5}$

:⟼  $(3 + 3) + ( \sqrt{5} - \sqrt{5} )$

:⟼  $9 + 0$

:⟼  $9$

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