Question

If the roots of the quadratic equation $x^2-px+q=0$ are real and differ by a quantity less than 1, then

• A. $b>\frac{a^2}{4}$
• B. $b<\frac{a^2-1}{4}$
• C. $\frac{a^2-1}{4}<b<\frac{a^2}{4}$
• D. None of these

Answer 1

The correct option is C $\frac{a^2-1}{4}<b<\frac{a^2}{4}$

Given equation
$x^2-px+q=0$
Comparing it with standard equation
$a{x}^2+bx+c=0$
$a=1,b=-p,c=q$
Given condition is roots of the given quadratic equation differs by a quantity less than $1$.
Mathematical representstion of this condition is
$|\alpha -\beta |<1$
We know that
$(\alpha -\beta {)}^2=\frac{D}{a^2}$

Now,
$\frac{\sqrt{D}}{|a|}<1$
$\Rightarrow \frac{\sqrt{(-p{)}^2-4\times 1\times q}}{1}<1$
$\Rightarrow \sqrt{p^2-4q}<1$
Squaring the inequality
$\Rightarrow |p^2-4q|<1$
$\Rightarrow -1<p^2-4q<1$
$\Rightarrow -1-p^2<-4q<1-p^2$
$\Rightarrow \frac{-1-p^2}{4}<-q<\frac{1-p^2}{4}$
$\Rightarrow \frac{p^2-1}{4}<q<\frac{1+p^2}{4}$