Question

If the roots of $x^2-px+q=0$ are two consecutive integers, find the value of $p^2-4q.$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Let the smaller roots be $n$

Then the other root is $n+1$

So, sum of roots$=n+(n+1)=-\frac{-p}{1}=p$

$\therefore 2n+1=p$

$\therefore n=\frac{p-1}{2}$ ...(1)

product of roots $=n(n+1)=\frac{q}{1}=q$

Substituting the value of $n$ from equation (1), we get

$\therefore \left(\frac{p-1}{2}\right)(\frac{p-1}{2}+1)=q$

$\therefore \left(\frac{p-1}{2}\right)\left(\frac{p+1}{2}\right)=q$

$\therefore \frac{p^2-1}{4}=q$

$\therefore p^2-4q=1$

So, the answer is option (A)