#### Question

If the equation x^{2} − bx + 1 = 0 does not possess real roots, then

##### Options

−3 < b < 3

−2 < b < 2

b > 2

b < −2

#### Solution

The given quadric equation is `x^2 - bx + 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 - 4ac`

Putting the value of a = 1, b = -b and ,c = 1

`=(-b)^2 - 4 xx 1 xx 1`

`=b^2 - 4`

The given equation does not have real roots, if D < 0

`b^2 - 4 < 0`

`b^2 < 4`

`b< sqrt4`

`b< ±2`

Therefore, the value of -2 < b< 2 .

Is there an error in this question or solution?

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If the Equation X2 − Bx + 1 = 0 Does Not Possess Real Roots, Then Concept: Solutions of Quadratic Equations by Factorization.

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