Question : If $x^2-2\sqrt{10}x+1=0$, what is the value of $(x-\frac{1}{x})$?
Option 1: $4$
Option 2: $6$
Option 3: $3$
Option 4: $5$
Correct Answer: $6$
Solution :
Given: $x^2-2\sqrt{10}x+1=0$
Dividing both sides by $x$, we get,
$⇒x-2\sqrt{10}+\frac{1}{x}=0$
$⇒x+\frac{1}{x}=2\sqrt{10}$
Squaring both sides, we get,
$⇒x^2+\frac{1}{x^2}+2×x×\frac{1}{x}=(2\sqrt{10})^2$
$⇒x^2+\frac{1}{x^2}=38$
Subtracting 2 from both sides, we get,
$⇒x^2+\frac{1}{x^2}-2=38-2$
$⇒x^2+\frac{1}{x^2}-2×x×\frac{1}{x}=36$
$⇒(x-\frac{1}{x})^2=36$
$\therefore x-\frac{1}{x}=\sqrt{36}=6$
Hence, the correct answer is $6$.
Related Questions
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
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