Verified Solution • 2025–26 Session moderate QJEEM23Maf1e1: JEE Main 2023 Mathematics - Class 11 JEEM23Maf1e1 Solution: Let be the roots of the equation and . Then is equal to… Let α1,α2,....,α7\alpha_1,\alpha_2,....,\alpha_7α1,α2,....,α7 be the roots of the equation x7+3x5−13x3−15x=0{x^7} + 3{x^5} - 13{x^3} - 15x = 0x7+3x5−13x3−15x=0 and ∣α1∣≥∣α2∣≥ ... ≥ ∣α7∣|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|∣α1∣≥∣α2∣≥...≥∣α7∣ . Then α1α2−α3α4+α5α6\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6α1α2−α3α4+α5α6 is equal to _________. Step-by-Step Verified Solution x7+3x5−13x3−15x=0{x^7} + 3{x^5} - 13{x^3} - 15x = 0x7+3x5−13x3−15x=0 x(x6+3x4−13x2−15)=0x({x^6} + 3{x^4} - 13{x^2} - 15) = 0x(x6+3x4−13x2−15)=0 x=0=α7x = 0 = {\alpha _7}x=0=α7 Let x2=t{x^2} = tx2=t t3+3t2−13t−15=0{t^3} + 3{t^2} - 13t - 15 = 0t3+3t2−13t−15=0 (t+1)(t+5)(t−3)=0(t + 1)(t + 5)(t - 3) = 0(t+1)(t+5)(t−3)=0 t=x2=−1,−5,3t = {x^2} = - 1, - 5,3t=x2=−1,−5,3 x = ± i,± 5i,± 3x\, = \, \pm \,i, \pm \,\sqrt 5 i, \pm \,\sqrt 3x=±i,±5i,±3 α1,α2=± 5i,α3,α4=± 3,α5,α6=± i{\alpha _1},{\alpha _2} = \pm \,\sqrt 5 i,{\alpha _3},{\alpha _4} = \pm \,\sqrt 3 ,{\alpha _5},{\alpha _6} = \pm \,iα1,α2=±5i,α3,α4=±3,α5,α6=±i α1α2−α3α4+α5α6=5+3+1=9{\alpha _1}{\alpha _2} - {\alpha _3}{\alpha _4} + {\alpha _5}{\alpha _6} = 5 + 3 + 1 = 9α1α2−α3α4+α5α6=5+3+1=9 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring