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Question

Find the value of positive integer ’ n’ for which the quadratic equation,nk=1(x+k1)(x+k)=10n, has solutions α and α+1 for some α
  1. 12
  2. 10
  3. 9
  4. 11


Solution

The correct option is D 11
nk=1(x+k1)(x+k)=10n[x2+(2k1)x+(k1)k]=10n
n.x2+n2x+(n1)n(n+1)310n=0
Let roots be α and β
α+β=n
αβ=(n1)(n+1)310
Since, difference of root is 1,
So,(α+β)24αβ=1
n24((n1)(n+1)3)=13n24(n2130)=33n24n2+124=3+n2=121n=±11n=11

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