If the equation x^2-bx+1=0 does not possess real roots then find value of b. Show method?

Answer 1

complete the square:

x^2-bx+1 = (x-(b/2))^2 -b^2/4 +1 = 0

(x-(b/2))^2 = b^2/4 -1

We need b^2/4 < 1 in order for anything to be imaginary when solving this, so:

b^2 < 4 or

|b| < 2

Answer 2

Use the quadratic eqn. Look at the radical, and if you can manipulate b such that there are no real roots, then...
x = {-(-b) +- sqr[(-b)^2 - 4(1)(1)]}/4(1)
= {b +- sqr[b^2 - 4]}/4
What would make x unreal?
when b^2-4 < 0 or b^2 < 2 or -2 < b < 2

Answer 3

This is of form ax^2+bx+c=0
b= -b
c=1
a=1
If no real roots, b^2-4ac < 0
i.e. (-b)^2-4 <0
b^2-4<0
b^2 < 4 ( all values of b with this property)

Answer 4

Assume b is real.
The roots are x = [ b +/- sqrt(b^2 - 4) ] / 2
If this is not real, then sqrt(b^2 - 4) is not real
Therefore, b^2 < 4
-2 < b < +2

If b is complex, then ...