X^2+(k-4)x+1=0 where k has no real roots, find the set of possible values of k

To solve for the values of \( k \) such that the quadratic equation \( x^2 + (k-4)x + 1 = 0 \) has no real roots, we need to analyze the **discriminant** of the quadratic equation.

For a general quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is given by:

\[
\Delta = b^2 - 4ac
\]

In your equation, \( a = 1 \), \( b = (k-4) \), and \( c = 1 \). Therefore, the discriminant is:

\[
\Delta = (k-4)^2 - 4(1)(1)
\]

Simplifying this expression:

\[
\Delta = (k-4)^2 - 4
\]

For the quadratic equation to have **no real roots**, the discriminant must be **less than zero**:

\[
(k-4)^2 - 4 < 0
\]

Now solve this inequality:

\[
(k-4)^2 < 4
\]

Taking the square root of both sides:

\[
|k-4| < 2
\]

This implies:

\[
-2 < k-4 < 2
\]

Adding 4 to all parts of the inequality:

\[
2 < k < 6
\]

Thus, the set of possible values of \( k \) such that the quadratic equation has no real roots is:

\[
k \in (2, 6)
\]

### Would you like further details or have any questions?

### Related questions:
1. What happens if the discriminant is equal to zero?
2. How would the roots change if \( k = 2 \)?
3. What if \( k > 6 \), how does the nature of the roots change?
4. Can you give a similar quadratic equation where the roots are imaginary?
5. How does completing the square help in finding the roots?

### Tip: Always check the discriminant when analyzing the nature of the roots of quadratic equations!

Discover how to solve the quadratic equation x^2 + (k-4)x + 1 = 0 by analyzing the discriminant to find the range of k that results in no real roots. This problem involves algebraic concepts such as the quadratic equation, discriminant, and inequality solving, ideal for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation