Question

If the difference of the roots of $x^2-px+q=0$ is unity, then

Answer 1

Let $\alpha ,\beta$ are the roots of $x^2-px+q=0$

Now,

$\text{sum of roots}=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha +\beta =p\dots \left(1\right)$

$\text{Product of roots}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha \beta =q\dots \left(2\right)$

Given,

$\mid \alpha -\beta \mid =1$

Squaring on both sides, we get

$\Rightarrow \mid \alpha -\beta {\mid }^2=1$

$\Rightarrow ({\alpha }^2+{\beta }^2-2\alpha \beta )=1$

$\Rightarrow ({\alpha }^2+{\beta }^2+2\alpha \beta )-4\alpha \beta =1$

$\Rightarrow (\alpha +\beta {)}^2-4\alpha \beta =1$

From $\left(1\right),\left(2\right)$, we get

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

Also, $p^2=1+4q$

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

$\Rightarrow p^2+4q^2=(1+2q{)}^2$

Hence, Option (B) and (C) both are correct.