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### Part (a): Show that [tex]\( k^2 - 4k - 12 > 0 \)[/tex].
To determine the conditions under which the quadratic equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex] has different real roots, we need to use the fact that the discriminant of the quadratic equation must be positive.
A quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has different real roots if and only if the discriminant [tex]\(\Delta\)[/tex] is greater than zero, where the discriminant is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the given equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = k \)[/tex]
- [tex]\( c = k + 3 \)[/tex]
The discriminant is:
[tex]\[ \Delta = k^2 - 4 \cdot 1 \cdot (k + 3) \][/tex]
[tex]\[ \Delta = k^2 - 4(k + 3) \][/tex]
[tex]\[ \Delta = k^2 - 4k - 12 \][/tex]
For the quadratic equation to have different real roots, the discriminant must be greater than zero:
[tex]\[ k^2 - 4k - 12 > 0 \][/tex]
Thus, we have shown that [tex]\( k^2 - 4k - 12 > 0 \)[/tex] is the condition for the quadratic equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex] to have different real roots.
### Part (b): Find the set of possible values of [tex]\( k \)[/tex].
To find the set of possible values of [tex]\( k \)[/tex] that satisfy [tex]\( k^2 - 4k - 12 > 0 \)[/tex], we need to solve the inequality.
Step 1: Solve the quadratic equation [tex]\( k^2 - 4k - 12 = 0 \)[/tex].
Factoring the quadratic equation:
[tex]\[ k^2 - 4k - 12 = (k - 6)(k + 2) = 0 \][/tex]
So, the roots are:
[tex]\[ k = 6 \quad \text{and} \quad k = -2 \][/tex]
Step 2: Determine the intervals where [tex]\( k^2 - 4k - 12 > 0 \)[/tex].
The roots divide the number line into three intervals:
1. [tex]\( (-\infty, -2) \)[/tex]
2. [tex]\( (-2, 6) \)[/tex]
3. [tex]\( (6, \infty) \)[/tex]
We need to test the sign of [tex]\( k^2 - 4k - 12 \)[/tex] in each of these intervals:
- For [tex]\( k \in (-\infty, -2) \)[/tex]:
Choose a test point, for example, [tex]\( k = -3 \)[/tex]:
[tex]\[ (-3)^2 - 4(-3) - 12 = 9 + 12 - 12 = 9 > 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 > 0 \)[/tex] in [tex]\((- \infty, -2)\)[/tex].
- For [tex]\( k \in (-2, 6) \)[/tex]:
Choose a test point, for example, [tex]\( k = 0 \)[/tex]:
[tex]\[ (0)^2 - 4(0) - 12 = -12 < 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 < 0 \)[/tex] in [tex]\((-2, 6)\)[/tex].
- For [tex]\( k \in (6, \infty) \)[/tex]:
Choose a test point, for example, [tex]\( k = 7 \)[/tex]:
[tex]\[ (7)^2 - 4(7) - 12 = 49 - 28 - 12 = 9 > 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 > 0 \)[/tex] in [tex]\( (6, \infty) \)[/tex].
Therefore, the inequality [tex]\( k^2 - 4k - 12 > 0 \)[/tex] is satisfied for:
[tex]\[ k \in (-\infty, -2) \cup (6, \infty) \][/tex]
So the set of possible values for [tex]\( k \)[/tex] is:
[tex]\[ (-\infty, -2) \cup (6, \infty) \][/tex].
This completes the solution.
### Part (a): Show that [tex]\( k^2 - 4k - 12 > 0 \)[/tex].
To determine the conditions under which the quadratic equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex] has different real roots, we need to use the fact that the discriminant of the quadratic equation must be positive.
A quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has different real roots if and only if the discriminant [tex]\(\Delta\)[/tex] is greater than zero, where the discriminant is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the given equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = k \)[/tex]
- [tex]\( c = k + 3 \)[/tex]
The discriminant is:
[tex]\[ \Delta = k^2 - 4 \cdot 1 \cdot (k + 3) \][/tex]
[tex]\[ \Delta = k^2 - 4(k + 3) \][/tex]
[tex]\[ \Delta = k^2 - 4k - 12 \][/tex]
For the quadratic equation to have different real roots, the discriminant must be greater than zero:
[tex]\[ k^2 - 4k - 12 > 0 \][/tex]
Thus, we have shown that [tex]\( k^2 - 4k - 12 > 0 \)[/tex] is the condition for the quadratic equation [tex]\( x^2 + kx + (k + 3) = 0 \)[/tex] to have different real roots.
### Part (b): Find the set of possible values of [tex]\( k \)[/tex].
To find the set of possible values of [tex]\( k \)[/tex] that satisfy [tex]\( k^2 - 4k - 12 > 0 \)[/tex], we need to solve the inequality.
Step 1: Solve the quadratic equation [tex]\( k^2 - 4k - 12 = 0 \)[/tex].
Factoring the quadratic equation:
[tex]\[ k^2 - 4k - 12 = (k - 6)(k + 2) = 0 \][/tex]
So, the roots are:
[tex]\[ k = 6 \quad \text{and} \quad k = -2 \][/tex]
Step 2: Determine the intervals where [tex]\( k^2 - 4k - 12 > 0 \)[/tex].
The roots divide the number line into three intervals:
1. [tex]\( (-\infty, -2) \)[/tex]
2. [tex]\( (-2, 6) \)[/tex]
3. [tex]\( (6, \infty) \)[/tex]
We need to test the sign of [tex]\( k^2 - 4k - 12 \)[/tex] in each of these intervals:
- For [tex]\( k \in (-\infty, -2) \)[/tex]:
Choose a test point, for example, [tex]\( k = -3 \)[/tex]:
[tex]\[ (-3)^2 - 4(-3) - 12 = 9 + 12 - 12 = 9 > 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 > 0 \)[/tex] in [tex]\((- \infty, -2)\)[/tex].
- For [tex]\( k \in (-2, 6) \)[/tex]:
Choose a test point, for example, [tex]\( k = 0 \)[/tex]:
[tex]\[ (0)^2 - 4(0) - 12 = -12 < 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 < 0 \)[/tex] in [tex]\((-2, 6)\)[/tex].
- For [tex]\( k \in (6, \infty) \)[/tex]:
Choose a test point, for example, [tex]\( k = 7 \)[/tex]:
[tex]\[ (7)^2 - 4(7) - 12 = 49 - 28 - 12 = 9 > 0 \][/tex]
So, [tex]\( k^2 - 4k - 12 > 0 \)[/tex] in [tex]\( (6, \infty) \)[/tex].
Therefore, the inequality [tex]\( k^2 - 4k - 12 > 0 \)[/tex] is satisfied for:
[tex]\[ k \in (-\infty, -2) \cup (6, \infty) \][/tex]
So the set of possible values for [tex]\( k \)[/tex] is:
[tex]\[ (-\infty, -2) \cup (6, \infty) \][/tex].
This completes the solution.
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