Question : If $x=1+\sqrt2+\sqrt3$, then find the value of $x^{2}-2x+4$.
Option 1: $2(7+\sqrt6)$
Option 2: $2(4+\sqrt6)$
Option 3: $2(3+\sqrt7)$
Option 4: $(4+\sqrt6)$
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Correct Answer: $2(4+\sqrt6)$
Solution : $x=1+\sqrt{2}+\sqrt{3}$ ⇒ $x-1=\sqrt{2}+\sqrt{3}$ Squaring both sides ⇒ $(x-1)^2=(\sqrt{2}+\sqrt{3})^2$ ⇒ $(x^2+1-2x)=(2+3+2\sqrt{6})$ ⇒ $(x^2+1-2x)=(5+2\sqrt{6})$ adding 3 to both sides, we get, ⇒ $(x^2+1-2x)+3=(5+2\sqrt{6})+3$ ⇒ $(x^2+4-2x)=(8+2\sqrt{6})$ $\therefore (x^2+4-2x)=2(4+\sqrt{6})$ Hence, the correct answer is $2(4+\sqrt{6})$.
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Question : If $x=\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}$, then what is the value of $(x^{2}+x-9)$?
Option 1: 0
Option 2: $3\sqrt2$
Option 3: $3\sqrt3$
Option 4: $5\sqrt3$
Question : If $x=\sqrt2+1$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Option 1: $8\sqrt2$
Option 2: $18\sqrt2$
Option 3: $6\sqrt2$
Option 4: $24\sqrt2$
Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{5}$
Question : When $2x+\frac{2}{x}=3$, then the value of ($x^3+\frac{1}{x^3}+2)$ is:
Option 1: $\frac{2}{7}$
Option 2: $\frac{7}{8}$
Option 3: $\frac{7}{2}$
Option 4: $\frac{8}{7}$
Question : If $(x^2-2x+1)=0$, then the value of $x^4+\frac{1}{x^4}$ is:
Option 2: 1
Option 3: 2
Option 4: 3
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