Question

plz do these 2 question on paper neatly so that i can understand​

Answer 1

Question: If x = 1 + √2 + √3, prove that x⁴ - 4x³ - 4x² + 16x - 8 = 0.

Solution:

$\sf x = 1 + \sqrt{2} + \sqrt{3} \\ \\ \sf \implies x - 1 = \sqrt{2} + \sqrt{3} \\ \\ \bf on \: squaring \: both \: sides - \\ \\ \sf \implies (x - 1) {}^{2} = ( \sqrt{2} + \sqrt{3} ) {}^{2} \\ \\ \sf \implies {x}^{2} - 2 \: . \: x \: . \: 1+ (1) {}^{2} = {( \sqrt{2} )}^{2} + ( \sqrt{3} ) {}^{2} + 2 \:. \: \sqrt{2} \: . \: \sqrt{3} \\ \\ \sf \implies x {}^{2} - 2x + 1 = 2 + 3 + 2 \sqrt{6} \\ \\ \sf \implies x {}^{2} - 2x + 1 = 5 + 2 \sqrt{6} \\ \\ \sf \implies x {}^{2} - 2x + 1 - 5 = 2 \sqrt{6} \\ \\ \sf \implies {x}^{2} - 2x - 4 = 2 \sqrt{6}$

On squaring both the sides-

$\sf \implies (x {}^{2} - 2x - 4) {}^{2} = {(2 \sqrt{6} )}^{2} \\ \\ \sf \implies (x {}^{2})^{2} + ( -2x) {}^{2} + ( - 4) {}^{2} + 2*x {}^{2} *( - 2x) \\\sf + 2*( - 2x)*( - 4) + 2*( - 4)*x {}^{2} = (2 \sqrt{6}) {}^{2} \\ \\\sf \implies x {}^{4} + 4x {}^{2} + 16 - 4x {}^{3} + 16x - 8x {}^{2} - 24 = 0 \\ \\ \boxed{\bf x {}^{4} -4x {}^{3} - 4x {}^{2} +16x- 8 = 0}$

Hence, it is proved.

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2. For second answer, refer to the attachment. Regrets for handwriting :)