Get the answers you've been searching for with IDNLearn.com. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
Certainly! Let's solve the given question step-by-step.
### 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.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
