The cubic equation which has three roots 1,3−√2i,3+√2i is
A
x3+7x2+17x−11=0
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B
x3−7x2+17x−11=0
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C
x3−7x2+17x+1=0
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D
x3+7x2−8=0
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Solution
The correct option is Bx3−7x2+17x−11=0 Roots are 1,3−√2i,3+√2i Sum of roots, S1=1+3−√2i+3+√2i=7 S2=1(3−√2i)+1(3+√2i)+(3−√2i)(3+√2i) =6+9+2=17 S3=1(3−√2i)(3+√2i)=9+2=11 Therefore equation is x3+(−7)x2+17x+(−11)=0⇒x3−7x2+17x−11=0