If the zeroes of the polynomial x2 bx+c are two consecutive positive integers, then b2 4ac

Text Solution

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Solution : Let the roots are `alpha` and `alpha+1`, where `alpha epsilonI` <br> Then sum of the roots `=2alpha+1=b` <br> Producrt of the roots `=alpha(alpha+1)=c` <br> Now `b^(2)-4c=(2alpha+1)^(2)-4alpha(alpha+1)` <br> `=4 alpha^(2)+1+4alpha-4alpha^(2)-4 alpha=1` <br> `:.b^(2)-4c=1`

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