If x = 1 + sqrt{2} + sqrt{3}, then find the v | vedclass

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(A) Given $x = 1 + \sqrt{2} + \sqrt{3}$.Rearranging the terms,we get $x - 1 = \sqrt{2} + \sqrt{3}$.Squaring both sides:$(x - 1)^{2} = (\sqrt{2} + \sqr

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$2(4 + \sqrt{6})$

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(A) Given $x = 1 + \sqrt{2} + \sqrt{3}$.Rearranging the terms,we get $x - 1 = \sqrt{2} + \sqrt{3}$.Squaring both sides:$(x - 1)^{2} = (\sqrt{2} + \sqrt{3})^{2}$$x^{2} - 2x + 1 = 2 + 3 + 2\sqrt{2 	imes 3}$$x^{2} - 2x + 1 = 5 + 2\sqrt{6}$$x^{2} - 2x = 4 + 2\sqrt{6}$Now,we need to find the value of $x^{2} - 2x + 4$.Substituting the value of $(x^{2} - 2x)$:$(4 + 2\sqrt{6}) + 4 = 8 + 2\sqrt{6}$$= 2(4 + \sqrt{6})$.

If $x = 1 + \sqrt{2} + \sqrt{3},$ then find the value of $x^{2} - 2x + 4$.

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