The equation whose roots are 1±2, 2±3 is

The Equation whose roots are 1±2, 2±3is (x−(1+2))(x−(1−2))(x−(2+3))(x−(2−3))=0⇒ (x2−2x−1)(x2−4x+1)=0⇒ x4−4x3+x2−2x3+8x2−2x−x2+4x−1=0⇒ x4−6x3+8x2+2x−1=0

The equation whose roots are 1±2, 2±3 is

The Equation whose roots are 1±2, 2±3is (x−(1+2))(x−(1−2))(x−(2+3))(x−(2−3))=0⇒ (x2−2x−1)(x2−4x+1)=0⇒ x4−4x3+x2−2x3+8x2−2x−x2+4x−1=0⇒ x4−6x3+8x2+2x−1=0

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The equation whose roots are $1 \pm \sqrt{2}, 2 \pm \sqrt{3}$ is

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$x^{4} - 5 x^{2} + 4 = 0$

$x^{4} - 9 x^{3} + 27 x^{2} - 33 x + 14 = 0$

$x^{4} - 6 x^{3} + 8 x^{2} + 2 x - 1 = 0$

$x^{4} - 6 x^{3} + 4 x^{2} + 54 x - 117 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} T h e E q u a t i o n w h o s e r o o t s a r e 1 \pm \sqrt{2}, 2 \pm \sqrt{3} \\ i s (x - (1 + \sqrt{2})) (x - (1 - \sqrt{2})) (x - (2 + \sqrt{3})) (x - (2 - \sqrt{3})) = 0 \\ \Rightarrow (x^{2} - 2 x - 1) (x^{2} - 4 x + 1) = 0 \\ \Rightarrow x^{4} - 4 x^{3} + x^{2} - 2 x^{3} + 8 x^{2} - 2 x - x^{2} + 4 x - 1 = 0 \\ \Rightarrow x^{4} - 6 x^{3} + 8 x^{2} + 2 x - 1 = 0 \end{array}$