Question

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

• A.
• B.
• C.
• D. none of these

Answer 1

Answer: (C)

$x^2-ax+b=0$
$\alpha +\beta =a$
$\alpha \beta =b$
$0<|\alpha -\beta |<1$
$0<\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }<1$
$0<\sqrt{a^2-4b}<1$
$0<a^2-4b<1$
$-1<4b-a^2<0$
$\therefore a^2-1<4b<a^2$
$\frac{a^2-1}{4}<b<\frac{a^2}{4}$