Question:
If the equation $x^{2}-b x+1=0$ does not possess real roots, then
(a) $-3 (b) $-2 (c) $b>2$ (d) $b<-2$
Solution:
The given quadric equation is $x^{2}-b x+1=0$, and does not have real roots.
Then find the value of b.
Here, $a=1, b=-b$ and,$c=1$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=1, b=-b$ and, $c=1$
$=(-b)^{2}-4 \times 1 \times 1$
$=b^{2}-4$
The given equation does not have real roots, if $D<0$
$b^{2}-4<0$
$b^{2}<4$
$b<\sqrt{4}$
$b<\pm 2$
Therefore, the value of $-2 Thus, the correct answer is $(b)$