To determine the range of possible values of [tex]\( k \)[/tex] such that the equation [tex]\( k x^2 + 6k x + 5 = 0 \)[/tex] has no real roots, let's follow these steps:
1. Understand the condition for no real roots:
For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] to have no real roots, its discriminant must be less than zero. The discriminant [tex]\( \Delta \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
2. Identify coefficients in the given quadratic equation:
The given equation is [tex]\( k x^2 + 6k x + 5 = 0 \)[/tex]. Here, the coefficients are:
[tex]\[
a = k, \quad b = 6k, \quad c = 5
\][/tex]
3. Compute the discriminant:
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[
\Delta = (6k)^2 - 4(k)(5)
\][/tex]
[tex]\[
\Delta = 36k^2 - 20k
\][/tex]
4. Set up the inequality for no real roots:
For the quadratic equation to have no real roots, the discriminant must be less than zero:
[tex]\[
36k^2 - 20k < 0
\][/tex]
5. Solve the inequality:
Factorize the quadratic expression:
[tex]\[
36k^2 - 20k = 4k(9k - 5)
\][/tex]
Hence, the inequality becomes:
[tex]\[
4k(9k - 5) < 0
\][/tex]
Since [tex]\( k \)[/tex] is a non-zero constant and must be positive:
[tex]\[
4k(9k - 5) < 0
\][/tex]
Analyze the factor [tex]\( 4k \)[/tex]:
- [tex]\( 4k \)[/tex] is always positive if [tex]\( k > 0 \)[/tex].
Analyze the factor [tex]\( (9k - 5) \)[/tex]:
- [tex]\( (9k - 5) < 0 \)[/tex] when [tex]\( 9k < 5 \)[/tex].
Therefore, the inequality [tex]\( 4k(9k - 5) < 0 \)[/tex] holds true when:
[tex]\[
9k - 5 < 0 \implies k < \frac{5}{9}
\][/tex]
6. Conclusion:
Combine the conditions found above. Since [tex]\( k \)[/tex] is a non-zero, positive constant:
[tex]\[
0 < k < \frac{5}{9}
\][/tex]
Thus, the range of possible values of [tex]\( k \)[/tex] such that the equation [tex]\( k x^2 + 6k x + 5 = 0 \)[/tex] has no real roots is:
[tex]\[
0 < k < \frac{5}{9}
\][/tex]
