Question : If $x+\frac{16}{x}=8$, then the value of $x^2+\frac{32}{x^2}$ is:
Option 1: 20
Option 2: 24
Option 3: 16
Option 4: 18
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Correct Answer: 18
Solution : Given: The value of $x+\frac{16}{x}=8$. We know the algebraic identity, $(a–b)^2=a^2+b^2–2ab$. $x+\frac{16}{x}=8$ ⇒ $x^2+16=8x$ ⇒ $x^2–8x+16=0$ ⇒ $(x–4)^2=0$ ⇒ $x=4,4$ The value of $x^2+\frac{32}{x^2}$ is given as, ⇒ $4^2+\frac{32}{4^2}=16+\frac{32}{16}$ = $16+2=18$ Hence, the correct answer is 18.
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Question : If $x^2-\frac{1}{x^2}=4 \sqrt{2}$, what is the value of $x^4-\frac{1}{x^4}?$
Option 1: $16 \sqrt{2}$
Option 2: $8\sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $32 \sqrt{2}$
Question : If $\frac{x}{2}-\frac{\left [4\left (\frac{15}{2}-\frac{x}{3} \right ) \right ]}{3} = –\frac{x}{18}$ then what is the value of $x$?
Option 1: –$10$
Option 2: $\frac{9}{8}$
Option 3: $10$
Option 4: $–\frac{9}{8}$
Question : If $(x + \frac{1}{x})$ = 6, then the value of ($x^{2} + \frac{1}{x^{2}}$) is:
Option 1: 23
Option 2: 16
Option 3: 34
Option 4: 32
Question : If $x^{2}-12x+33=0$, then what is the value of $(x-4)^{2}+\frac{1}{(x-4)^{2}}?$
Option 1: $16$
Option 2: $14$
Option 3: $18$
Option 4: $20$
Question : If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to:
Option 1: 18
Option 2: 12
Option 3: 15
Option 4: 16
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