Question

if a=7-4√3, find the value of √a+1/√a

Answer 1

your answer is here ✌

Answer 2

The value of $\bold{\sqrt{a}+\frac{1}{\sqrt{a}}}$ is 4

Given:

$a=7-4 \sqrt{3}$

To find:

The value of $\sqrt{a}+\frac{1}{\sqrt{a}}$

Solution:

The given value of a is $7-4 \sqrt{3}$

And the value of $\frac{1}{a} \text { is } 7+4 \sqrt{3}$

Then,  

$\begin{array}{c}{a+\frac{1}{a}=7-4 \sqrt{3}+7+4 \sqrt{3}} \\ {a+\frac{1}{a}=14}\end{array}$

Now to find the value of $\sqrt{a}+\frac{1}{\sqrt{a}}$,

By using the formula of $(a+b)^{2}=a^{2}+b^{2}+2 a b$

Then, applying the $(a+b)^{2}$ formula to $\sqrt{a}+\frac{1}{\sqrt{a}}$

We can get,

$\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=(\sqrt{a})^{2}+\left(\frac{1}{\sqrt{a}}\right)^{2}+\left(2 \times \sqrt{a} \times \frac{1}{\sqrt{a}}\right)} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=a+\frac{1}{a}+2}\end{array}$

Already we know the value of  $a+\frac{1}{a}$ is 14

$\begin{array}{c}{\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=14+2} \\ {\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}=16}\end{array}$

By taking square root for the value of $\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^{2}$, we can get the value of $\sqrt{a}+\frac{1}{\sqrt{a}}$

$\begin{array}{l}{\sqrt{a}+\frac{1}{\sqrt{a}}=\sqrt{16}} \\ {\sqrt{a}+\frac{1}{\sqrt{a}}=4}\end{array}$

Then the value of $\bold{\sqrt{a}+\frac{1}{\sqrt{a}}}$ is 4, if the value of a is  $7-4 \sqrt{3}$.