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To determine the range of possible values of [tex]\( k \)[/tex] such that the quadratic equation [tex]\( k x^2 + 6 k x + 5 = 0 \)[/tex] has no real roots, we need to use the discriminant of the quadratic equation. For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( k x^2 + 6 k x + 5 = 0 \)[/tex]:
- [tex]\( a = k \)[/tex]
- [tex]\( b = 6k \)[/tex]
- [tex]\( c = 5 \)[/tex]
The discriminant [tex]\(\Delta\)[/tex] for this quadratic equation is:
[tex]\[ \Delta = (6k)^2 - 4(k)(5) \][/tex]
Simplifying the discriminant:
[tex]\[ \Delta = 36k^2 - 20k \][/tex]
For the quadratic equation to have no real roots, the discriminant must be less than zero:
[tex]\[ \Delta < 0 \implies 36k^2 - 20k < 0 \][/tex]
To solve this inequality, we first factor the quadratic expression:
[tex]\[ 36k^2 - 20k = 4k(9k - 5) \][/tex]
So, the inequality becomes:
[tex]\[ 4k(9k - 5) < 0 \][/tex]
We need to determine the values of [tex]\( k \)[/tex] for which this product is negative. We start by finding the roots of the equation [tex]\( 4k(9k - 5) = 0 \)[/tex]:
[tex]\[ 4k = 0 \quad \text{or} \quad 9k - 5 = 0 \][/tex]
Solving these, we get:
[tex]\[ k = 0 \quad \text{or} \quad k = \frac{5}{9} \][/tex]
Since [tex]\( k \neq 0 \)[/tex] (as given in the problem), we discard [tex]\( k = 0 \)[/tex], leaving [tex]\( k = \frac{5}{9} \)[/tex].
Next, we analyze the sign of the expression [tex]\( 4k(9k - 5) \)[/tex] around the root [tex]\( k = \frac{5}{9} \)[/tex]. To do this, we test the intervals divided by [tex]\( k = \frac{5}{9} \)[/tex]:
1. Interval [tex]\( (-\infty, \frac{5}{9}) \)[/tex]: Choose [tex]\( k = 0 \)[/tex] (as in this interval but [tex]\( k \neq 0 \)[/tex]).
[tex]\[ 4 \cdot 0 \cdot (9 \cdot 0 - 5) = -0 < 0 \quad \text{(negative)} \][/tex]
2. Interval [tex]\( (\frac{5}{9}, \infty) \)[/tex]: Choose [tex]\( k = 1 \)[/tex] (as in this interval).
[tex]\[ 4 \cdot 1 \cdot (9 \cdot 1 - 5) = 4 \cdot 1 \cdot 4 = 16 > 0 \quad \text{(positive)} \][/tex]
Since [tex]\( 4k(9k - 5) < 0 \)[/tex] in the interval [tex]\( (-\infty, \frac{5}{9}) \)[/tex], and [tex]\( k \neq 0 \)[/tex], the range of possible values of [tex]\( k \)[/tex] for which the quadratic equation has no real roots is:
[tex]\[ \boxed{0 < k < \frac{5}{9}} \][/tex]
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( k x^2 + 6 k x + 5 = 0 \)[/tex]:
- [tex]\( a = k \)[/tex]
- [tex]\( b = 6k \)[/tex]
- [tex]\( c = 5 \)[/tex]
The discriminant [tex]\(\Delta\)[/tex] for this quadratic equation is:
[tex]\[ \Delta = (6k)^2 - 4(k)(5) \][/tex]
Simplifying the discriminant:
[tex]\[ \Delta = 36k^2 - 20k \][/tex]
For the quadratic equation to have no real roots, the discriminant must be less than zero:
[tex]\[ \Delta < 0 \implies 36k^2 - 20k < 0 \][/tex]
To solve this inequality, we first factor the quadratic expression:
[tex]\[ 36k^2 - 20k = 4k(9k - 5) \][/tex]
So, the inequality becomes:
[tex]\[ 4k(9k - 5) < 0 \][/tex]
We need to determine the values of [tex]\( k \)[/tex] for which this product is negative. We start by finding the roots of the equation [tex]\( 4k(9k - 5) = 0 \)[/tex]:
[tex]\[ 4k = 0 \quad \text{or} \quad 9k - 5 = 0 \][/tex]
Solving these, we get:
[tex]\[ k = 0 \quad \text{or} \quad k = \frac{5}{9} \][/tex]
Since [tex]\( k \neq 0 \)[/tex] (as given in the problem), we discard [tex]\( k = 0 \)[/tex], leaving [tex]\( k = \frac{5}{9} \)[/tex].
Next, we analyze the sign of the expression [tex]\( 4k(9k - 5) \)[/tex] around the root [tex]\( k = \frac{5}{9} \)[/tex]. To do this, we test the intervals divided by [tex]\( k = \frac{5}{9} \)[/tex]:
1. Interval [tex]\( (-\infty, \frac{5}{9}) \)[/tex]: Choose [tex]\( k = 0 \)[/tex] (as in this interval but [tex]\( k \neq 0 \)[/tex]).
[tex]\[ 4 \cdot 0 \cdot (9 \cdot 0 - 5) = -0 < 0 \quad \text{(negative)} \][/tex]
2. Interval [tex]\( (\frac{5}{9}, \infty) \)[/tex]: Choose [tex]\( k = 1 \)[/tex] (as in this interval).
[tex]\[ 4 \cdot 1 \cdot (9 \cdot 1 - 5) = 4 \cdot 1 \cdot 4 = 16 > 0 \quad \text{(positive)} \][/tex]
Since [tex]\( 4k(9k - 5) < 0 \)[/tex] in the interval [tex]\( (-\infty, \frac{5}{9}) \)[/tex], and [tex]\( k \neq 0 \)[/tex], the range of possible values of [tex]\( k \)[/tex] for which the quadratic equation has no real roots is:
[tex]\[ \boxed{0 < k < \frac{5}{9}} \][/tex]
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