MCQ
If $\text{a}=7-4\sqrt{3},$ then the value of $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}$ is:
- A$8$
- B$1$
- C$2$
- ✓$4$
✓
Answer
Correct option: D.
$4$
Let $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=\text{x}$
Then, squaring both side, we get
$\text{a}+\frac{1}{\text{a}}+2=\text{x}^2$
$\Rightarrow\frac{\text{a}^2+1}{\text{a}}+2=\text{x}^2$
Now, put the value of $a,$
$\frac{(7-4\sqrt{3})^2+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{49+48-56\sqrt{3}+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{98-56\sqrt{3}}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow14\Big(\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\Big)+2=\text{x}^2$
$\Rightarrow16=\text{x}^2$
$\Rightarrow\text{x}=4$
So, $\text{x}=\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=4$
Then, squaring both side, we get
$\text{a}+\frac{1}{\text{a}}+2=\text{x}^2$
$\Rightarrow\frac{\text{a}^2+1}{\text{a}}+2=\text{x}^2$
Now, put the value of $a,$
$\frac{(7-4\sqrt{3})^2+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{49+48-56\sqrt{3}+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{98-56\sqrt{3}}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow14\Big(\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\Big)+2=\text{x}^2$
$\Rightarrow16=\text{x}^2$
$\Rightarrow\text{x}=4$
So, $\text{x}=\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=4$
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