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Question: If n and r are positive integers such that 0 \< r \< n, then the roots of the quadratic equation <su...

If n and r are positive integers such that 0 < r < n, then the roots of the quadratic equation nCr – 1 x2 + 2 . nCrx + nCr + 1 = 0 are –

A

Real and distinct

B

Rational

C

Rational but not integer

D

Imaginary

Answer

Real and distinct

Explanation

Solution

The discriminant of the given equation is

D = 4((nCr)2−nCr−1nCr+1)\left( (^{n}C_{r})^{2} -^{n}{C_{r - 1}}^{n}C_{r + 1} \right)

= 4(a – b), where a = (nCr)2, b = nCr – 1. nCr + 1

Now, ab\frac{a}{b}=nCr.nCrnCr−1.nCr+1\frac{nC_{r}.^{n}C_{r}}{nC_{r - 1}.^{n}C_{r + 1}}

= r+1r\frac{r + 1}{r}. n−r+1n−r\frac{n - r + 1}{n - r}= (1+1r)\left( 1 + \frac{1}{r} \right) (1+1n−r)\left( 1 + \frac{1}{n - r} \right)> 1

\a > b Ž D > 0

Ž roots of given equation are real and distinct.