Question

If the equation x² − bx + 1 = 0 does not possess real roots, then
(a) −3 < b < 3
(b) −2 < b < 2
(c)b > 2
(d)b < −2

Answer 1

SOLUTION :  

Option (b) is correct :   - 2 < b < 2

Given : x² - bx + 1 = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1, b = - b  , c = 1

D(discriminant) = b² – 4ac

D = (- b)² - 4 × 1 × 1

D = b² - 4

D < 0 (Given : not real roots)

b² - 4 < 0

b² < 4

b < √4

b < ± 2

The value is  - 2 < b < 2

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Solution :

Compare given Quadratic

equation x² - bx + 1 = 0 with

Ax² + Bx + C = 0 , ,we get

A = 1 , B = - b , C = 1

Now,

Discreminant ( D ) <0

[ Given , equation has no

real roots ]

( -b )² - 4×1×1 < 0

=> b² - 4 < 0

=> b² < 4

=> b < ± 2

Therefore,

-2 < b < 2

Option ( b ) is correct.

•••••