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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