Question

11. If the equation x^2- 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

Answer:

If the eqn does not posses real root, then discriminant D must be less than zero

D = (-b)^2 - 4×1×1 must be less than 0

b^2 - 4 must be less than 0

b^2 must be less than 4

b must be less than 2 and -2

Answer 2

Answer:

d) b<-2

Step-by-step explanation:

we know that D=b²-4ac

and if D=0,then roots are real and equal

if D>0 ,then roots are real and unequal

if D<0 then roots are not real

and here we see that

A=1(co efficient of x²)

B= -b(co efficient of x)

C=1(constant term)

(-B)²-4*1*1 <0

B²-4 <0

{note that you can't transfer 4 from RHS to LHS or LHS to RHS because the sign between RHS and LHS is (<) ,not LHS equal to RHS}

adding 4 both side ( adding same number both side can't change their value)

B²-4+4<0+4

B²<4

B<√4

B<± 2

so

B<-2

so option D is correct