Question

If the roots of the equation $x^2-nx+m=0$ differ by 1, then

• A. $n^2-4m-1=0$
• B. $n^2+4m-1=0$
• C. $m^2+4n+1=0$
• D. $m^2-4n-1=0$

Answer 1

The correct option is A $n^2-4m-1=0$

Let $\alpha$ be one root and $\alpha +1$ be another root of the equation.

So,

$\alpha +\alpha +1=n\Rightarrow 2\alpha +1=n$

or, $\alpha =\frac{n-1}{2}$

And, $\alpha (\alpha +1)=m$

or, $\left(\frac{n-1}{2}\right)(\frac{n-1}{2}+1)=m$

or, $(n-1)(n+1)=4m$

or, $n^2-4m-1=0$

Hence, A is the correct option.