InterviewSolution
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InterviewSolution
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If, for a positive integer `n ,`the quadratic equation,`x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n`has two consecutive integral solutions, then `n`is equal to :`10`(2) `11`(3) `12`(4) `9`A. 12B. 9C. 10D. 11 |
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Answer» Correct Answer - D Given quadratic equation is `x(x+1)+(x+1)(x+2)+...+(x+bar(n-1))(x+n)=10m` `implies(x^(2)+x^(2)+..+x^(2))+[(1+3+5+...+(2n-1)]x+[(1.2+2.3+...+(n-1)n]=10n` `impliesnx^(2)+nx+(n(n^(2)-1))/(3)-10n=0` `x^(2)+nx+(n^(2)-1)/(3)-10=0` `implies3x^(2)+3nx+n^(2)-31=0` Let `alpha and beta` be the roots. Since, `alpha and beta` are consective. `therefore|alpha-beta|=1implies(alpha-beta)^(2)=1` Again, `(alpha-beta)^(2)=(alpha+beta)^(2)-4alpha beta` `implies1=((-3n)/(3))^(2)-4((n^(2)-31)/(3))` `implies1=n^(2)-4/3(n^(2)-31)implies3=3n^(2)-4n^(2)+124` `impliesn^(2)=121=n=+-11` `therefore" "n=11" "[becausengt0]` |
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