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If the roots of x^2-b x+c=0 are two cons...

If the roots of `x^2-b x+c=0` are two consecutive integers, then `b^2-4c` is `0` (b) `1` (c) `2` (d) none of these

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
Let the roots are `alpha` and `alpha+1`, where `alpha epsilonI`
Then sum of the roots `=2alpha+1=b`
Producrt of the roots `=alpha(alpha+1)=c`
Now `b^(2)-4c=(2alpha+1)^(2)-4alpha(alpha+1)`
`=4 alpha^(2)+1+4alpha-4alpha^(2)-4 alpha=1`
`:.b^(2)-4c=1`
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