Question

State the following statement is True or False

If the roots of the equation $x^2+px+q=0$ differ by $1$, then $p^2=1+4q$

• A. True
• B. False

Answer 1

The correct option is A True

The equation is : $x^2+px+q=0$
The roots of the equation differ by $1$

$\therefore x=\frac{-p\pm \sqrt{p^2-4q}}{2}$

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

$-p+\sqrt{p^2-4q}+p+\sqrt{p^2-4q}=2$

$2\sqrt{p^2-4q}=2$

$p^2-4q=1$

$p^2=1+4q$