MCQ
If $\text{x}=\sqrt6+\sqrt5,$ then $\text{x}^2+\frac{1}{\text{x}^2}-2=$
- A$2\sqrt6$
- B$2\sqrt5$
- C$24$
- ✓$20$
✓
Answer
Correct option: D.
$20$
$\text{x}^2+\frac{1}{\text{x}^2}-2=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2$
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}$
$ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
Now,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\big[\sqrt6+\sqrt5-\big(\sqrt6-\sqrt5\big]^2$
$\ =\big(2\sqrt5\big)^2=4\times5=20$
Hence, correct option is $(d)$.
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}$
$ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
Now,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\big[\sqrt6+\sqrt5-\big(\sqrt6-\sqrt5\big]^2$
$\ =\big(2\sqrt5\big)^2=4\times5=20$
Hence, correct option is $(d)$.
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