Question

If the roots of the equation $x^2-px+q=0$ differ by unity, then.

• A. $p^2=4q$
• B. $p^2=4q+1$
• C. $p^2=4q-1$
• D. None of these

Answer 1

Answer: (B)

$p^2=4q+1$

Let $\alpha$ and $\alpha +1$ be roots of $x^2-px+q=0$

Then, $\alpha +(\alpha +1)=-(-p)$

$\Rightarrow 2\alpha +1=p$

$\Rightarrow \alpha =\frac{p-1}{2}$ ....... $\left(i\right)$

Also, $\alpha (\alpha +1)=q$

$\Rightarrow {\alpha }^2+\alpha =q$ ......... $\left(ii\right)$

From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get

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

[on eliminating $\alpha$]

$\Rightarrow \left(\frac{p^2-2p+1}{4}\right)+\left(\frac{p-1}{2}\right)=q$

$\Rightarrow p^2-2p+1+2p-2=4q$

$\Rightarrow p^2=4q+1$.