If the equation x2 − bx + 1 = 0 does not possess real roots, then
Question:

If the equation $x^{2}-b x+1=0$ does not possess real roots, then

(a) $-3<b<3$

(b) $-2<b<2$

(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 $-2<b<2$

Thus, the correct answer is $(b)$

 

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