sum of reciprocals of the roots

Level pending

if L,M ,N are the roots of the equation ( x^3 + 2(x^2) -3(x) - 1 = 0) , then what is the sum of (1/L)^2 + ( 1/M)^2 + (1/N)^2 ????


The answer is 13.

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1 solution

Deepak Bharadwaj
Jan 17, 2014

L+M+N= -2(call it as equation 1) LM+MN+NL = -3(call it as equation 2) LMN=1(call it as equation 3)

square the equation 2 (LM+MN+NL)^2=(-3)^2
[(L)^2(M)^2]+[(M)^2(N)^2]+[(N)^2(L)^2] +2(LM+MN+NL)=9 { ((a+b+c)^2=a^2+b^2+c^2 +2(ab+bc+ca))} [(L)^2(M)^2]+[(M)^2(N)^2]+[(N)^2(L)^2] +2(-3)=9 [(L)^2(M)^2]+[(M)^2(N)^2]+[(N)^2(L)^2] =15

([(L)^2(M)^2]+[(M)^2(N)^2]+[(N)^2(L)^2])/((LMN)^2)=15/1

[([(L)^2(M)^2])/((L^2)(M^2)(N^2))] +[([[(M)^2(N)^2])/((L^2)(M^2)(N^2))]+[([(N)^2(L)^2])/((L^2)(M^2)(N^2))]=15

therefore,L^-2+M^-2+N^-2=15

The right answer is 13. There is a mistake in the solution given. When the equation ( L M + M N + N L ) 2 = ( 3 ) 2 (LM+MN+NL)^2=(-3)^2 is squared the result is L 2 M 2 + M 2 N 2 + N 2 L 2 + 2 ( L M L N + L M M N + M N N L ) = 9 L^2M^2+M^2N^2+N^2L^2 +2(LMLN+LMMN+MNNL)=9 instead of L 2 M 2 + M 2 N 2 + N 2 L 2 + 2 ( L M + M N + N L ) = 9 L^2M^2+M^2N^2+N^2L^2 +2(LM+MN+NL)=9 Therefore L 2 M 2 + M 2 N 2 + N 2 L 2 = 9 2 ( L M L N + L M M N + M N N L ) = L^2M^2+M^2N^2+N^2L^2 = 9 - 2(LMLN+LMMN+MNNL) = = 9 2 L M N ( L + M + N ) = 9 2 1 ( 2 ) = 13 = 9 - 2LMN(L+M+N) = 9 - 2*1*(-2) = 13

Later the solution follows as the one given. The answer was also verified using Wolfram Alpha.

Marin Shalamanov - 7 years, 1 month ago

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Thanks! I've updated the answer.

Calvin Lin Staff - 7 years, 1 month ago

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