Question : If $x^2-7x+1=0$, what is the value of $(x+\frac{1}{x})$.
Option 1: 7
Option 2: 3
Option 3: 51
Option 4: 47
Correct Answer: 7
Solution : Given: $x^2-7x+1=0$ Dividing both sides by $x$ we get, $x-7+\frac{1}{x}=0$ $\therefore x+\frac{1}{x}=7$ Hence, the correct answer is 7.
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Question : If $x^2-3x+1=0$, what is the value of $(x^4+\frac{1}{x^4})$?
Option 1: $11$
Option 2: $18$
Option 3: $47$
Option 4: $51$
Question : If $x^4+x^{-4}=47, x>0$, then what is the value of $x+\frac{1}{x}-2?$
Option 1: 1
Option 2: 0
Option 3: 5
Option 4: 3
Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
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