MathematicsEasy53×since 2002Q3243If (3+i)100=299(p+iq){\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)(3+i)100=299(p+iq), then p and q are roots of the equation :Ax2−(3−1)x−3=0{x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0x2−(3−1)x−3=0Bx2+(3+1)x+3=0{x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0x2+(3+1)x+3=0Cx2+(3−1)x−3=0{x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0x2+(3−1)x−3=0Dx2−(3+1)x+3=0{x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0x2−(3+1)x+3=0Check answerSkip