Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 162367 by cortano last updated on 29/Dec/21

Let x_1 ,x_2 ,x_3 be the roots of the equation x^3 +3x+5=0 . Then the value of expression (x_1 +(1/x_1 ))(x_2 +(1/x_2 ))(x_3 +(1/x_3 )) is equal to

Letx1,x2,x3betherootsoftheequationx3+3x+5=0.Thenthevalueofexpression(x1+1x1)(x2+1x2)(x3+1x3)isequalto

Answered by mr W last updated on 29/Dec/21

x_1 +x_2 +x_3 =0 x_1 x_2 +x_2 x_3 +x_3 x_1 =3 x_1 x_2 x_3 =−5 (x_1 +(1/x_1 ))(x_2 +(1/x_2 ))(x_3 +(1/x_3 )) =(((x_1 ^2 +1)(x_2 ^2 +1)(x_3 ^2 +1))/(x_1 x_2 x_3 )) =(((x_1 x_2 x_3 )^2 +(x_1 ^2 x_2 ^2 +x_2 ^2 x_3 ^2 +x_3 ^2 x_1 ^2 )+(x_1 ^2 +x_2 ^2 +x_3 ^2 )+1)/(x_1 x_2 x_3 )) =(((x_1 x_2 x_3 )^2 +(x_1 x_2 +x_2 x_3 +x_3 x_1 )^2 −2x_1 x_2 x_3 (x_1 +x_2 +x_3 )+(x_1 +x_2 +x_3 )^2 −2(x_1 x_2 +x_2 x_3 +x_3 x_1 )+1)/(x_1 x_2 x_3 )) =(((−5)^2 +(3)^2 −2(−5)(0)+(0)^2 −2(3)+1)/(−5)) =((25+9−6+1)/(−5)) =−((29)/5)

x1+x2+x3=0x1x2+x2x3+x3x1=3x1x2x3=−5(x1+1x1)(x2+1x2)(x3+1x3)=(x12+1)(x22+1)(x32+1)x1x2x3=(x1x2x3)2+(x12x22+x22x32+x32x12)+(x12+x22+x32)+1x1x2x3=(x1x2x3)2+(x1x2+x2x3+x3x1)2−2x1x2x3(x1+x2+x3)+(x1+x2+x3)2−2(x1x2+x2x3+x3x1)+1x1x2x3=(−5)2+(3)2−2(−5)(0)+(0)2−2(3)+1−5=25+9−6+1−5=−295

Terms of Service

Privacy Policy

Contact: info@tinkutara.com