# What are the values of b for which the equation 16x^2 - (b^2 - b)x + 12 has real roots.

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### 1 Answer

The values of b for which the quadratic equation 16x^2 - (b^2 - b)x + 12 has real roots have to be determined.

A quadratic equation ax^2 + bx + c = 0 has real roots if b^2 >= 4*a*c. The given equation has real roots if:

`(b^2 - b)^2 >= 768`

=> `b^2 - b >= sqrt 768` and `b^2 - b <= -sqrt 768`

=> `b^2 - b - sqrt 768 >= 0` and `b^2 - b + sqrt 768 <= 0`

`b^2 - b - sqrt 768 >= 0`

=> `(b + (sqrt(64*sqrt(3)+1)-1)/2)(b - (sqrt(64*sqrt(3)+1)+1)/2 )>= 0`

This is true for `b >= (sqrt(64*sqrt(3)+1)+1)/2)` and b `<= -(sqrt(64*sqrt(3)+1)+1)/2 `

`b^2 - b <= -sqrt 768`

=> `b^2 - b + sqrt 768 <= 0`

=> `(b-(sqrt(1-64*sqrt(3))-1)/2)(b-(sqrt(1-64*sqrt(3))+1)/2 )<=0`

This is true when b lies betweenÂ `(sqrt(1-64*sqrt(3))-1)/2` and `(sqrt(1-64*sqrt(3))+1)/2` .