182k views
2 votes
Show that x 2 - 6x + 11 > 0 for all real values of x

1 Answer

4 votes

Final answer:

To show that x^2 - 6x + 11 > 0 for all real values of x, we can use the discriminant of a quadratic equation.

Step-by-step explanation:

To show that x^2 - 6x + 11 > 0 for all real values of x, we can use the discriminant of a quadratic equation. The discriminant is the part of the quadratic formula that determines the number of solutions. In this case, the discriminant is b^2 - 4ac. For our equation, a = 1, b = -6, and c = 11.

Substituting these values into the discriminant, we get: (-6)^2 - 4(1)(11) = 36 - 44 = -8. Since the discriminant is negative, it means that the quadratic equation has no real solutions. Therefore, x^2 - 6x + 11 > 0 for all real values of x.

User Martin Solac
by
7.7k points

No related questions found