Question

If difference of roots of the equation $x^2-px+q=0$ is $1$, then $p^2+{4q}^2$ equals

• A. $2q+3$
• B. ${(1-2q)}^2$
• C. ${(1+2q)}^2$
• D. $2q-3$

Answer 1

The correct option is C ${(1+2q)}^2$

$x^2-px+q=0$

Let the root of the quadratic equation be $\alpha ,\beta$

$\therefore \alpha -\beta =1$

$(\alpha -\beta {)}^2=1$ (squaring both sides)

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

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

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

Sum of roots $\alpha$ and $\beta \Rightarrow \alpha +\beta =p$ and,

product of roots $\alpha$ and $\beta \Rightarrow \alpha \beta =q$

Substituting there values in equation (1), we get :

$p^2-4q=1$

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

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

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