From minus to plus

$\large{\begin{aligned}\left(a+\dfrac{1}{a}\right)^2 &= a^2+2a\left(\dfrac{1}{a}\right) + \dfrac{1}{a^2} \\ &= a^2+\dfrac{1}{a^2}+2 \\ &= a^2+\dfrac{1}{a^2} -2 + 4 \\ &= a^2 - 2a\left(\dfrac{1}{a}\right) + \dfrac{1}{a^2} + 4 \\ &= \left(a-\dfrac{1}{a}\right)^2 + 4 \\ &= \left(\dfrac{5}{6}\right)^2 + 4 \\ &= \dfrac{25}{36} + 4 \\\Rightarrow \left(a+\dfrac{1}{a}\right)^2 &= \dfrac{169}{36} \\ \Rightarrow \sqrt{\left(a+\dfrac{1}{a}\right)^2} &= \sqrt{\dfrac{169}{36}} \\\Rightarrow \left| a+\dfrac{1}{a} \right| &= \left| \dfrac{13}{6} \right| \\ \Rightarrow a+\dfrac{1}{a} &= \boxed{\dfrac{13}{6}} \dots \text{since a > 0} \end{aligned}}$

We have $( a + \dfrac{1}{a} )^{2} = ( a - \dfrac{1}{a} )^2 + 4 = (\dfrac{5}{6})^2 + 4 = \dfrac{169}{36}$ . But as $a > 0 \Rightarrow ( a + \dfrac{1}{a} ) > 0$ , we have $( a + \dfrac{1}{a} ) = + \sqrt{\dfrac{169}{36}} = \boxed{\dfrac{13}{6}}$ .

The options are poorly drafted.. Anyone who knows that sum of a number and its recirpocal is always >=2 will straight away mark 13/6 because that is the only option >=2

$a - 1/a = 5/6$

Squaring both sides and adding 4 both sides :

$(a + 1/a)^2 = 169/36$

so a + 1/a = 13/6

I hope that helps :)

a-1/a=5/6; 6a^2-5a-6=0; a1=3/2, bacause a>0; a+1/a=3/2+2/3=13/6

a - 1/a = 5/6 Therefore, a^2/a - 1/a = 5/6 (a^2 -1)/a = 5/6 6(a^2-1) = 5a (cross multiply) 6a^2 - 6 = 5a 6a^2 - 5a - 6 =0 (3a+2)(2a-3) = 0 a = -2/3 or a = 3/2 Since a > 0, a = 3/2 Substitute a into a +1/a 3/2 + 1/(3/2) = ? 3/2 + 2/3 = ? 9/6 + 4/6 = 13/6 =13/6

a-1/a= 5/6

6a^2 - 5a - 6=0

( 3a+2) (2a-3)= 0

But a>0

a = 3/2

So a+1/a= 3/2 +2/3

= 13/ 6

a-1/a=5/6 6a^2-6=5a 6a^2-5a-6=0 3a(2a-3)+2(2a-3)=0 a=-2/3,3/2 as a>0 hence 3/2+1/3/2=3/2+2/3 =13/6

a-\frac{1}{3}=\frac{5}{6} /*6a 6a^{2}-5a-6=0 (2a-3)(3a+2)=0 The only possible solution for a>0 is a=\frac{3}{2} \frac{3}{2}+1/\frac{3}{2}=\frac{13}{6}

a - 1/a = 5/6 es una cuadrática (6a^2 - 5a - 6 = 0), entonces tiene dos valores para "a" que la satisfasen (a = 3/2 y a = -2/3). Por lo tanto: a + 1/a = (+ ó -) 13/6. Esta es la respuesta. La condición a > 0 no se justifica, en tal caso la solución es +13/6

You dont have to that much of calculation a+1/a >2 as am>gm of two numbers so only last option satisfy given condition.