Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Certainly! Let's solve the system of equations step-by-step to determine the values of [tex]\( k \)[/tex] where the system has no real solutions, one real solution, and two real solutions.
### Given System of Equations:
[tex]\[ \begin{cases} y = x^2 \\ y = x + k \end{cases} \][/tex]
### Step 1: Set the equations equal to each other
Since both equations are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ x^2 = x + k \][/tex]
### Step 2: Rearrange into standard quadratic form
Rearrange the equation to the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ x^2 - x - k = 0 \][/tex]
### Step 3: Determine the discriminant
The discriminant of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For our equation [tex]\(x^2 - x - k = 0\)[/tex], we have [tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -k\)[/tex]. Substituting these values into the formula for the discriminant, we get:
[tex]\[ \Delta = (-1)^2 - 4(1)(-k) = 1 + 4k \][/tex]
### Step 4: Analyze the discriminant
The number of real solutions to the quadratic equation depends on the value of the discriminant [tex]\(\Delta\)[/tex]:
- No real solutions: [tex]\(\Delta < 0\)[/tex]
- One real solution: [tex]\(\Delta = 0\)[/tex]
- Two real solutions: [tex]\(\Delta > 0\)[/tex]
#### For no real solutions: [tex]\(\Delta < 0\)[/tex]
[tex]\[ 1 + 4k < 0 \][/tex]
[tex]\[ 4k < -1 \][/tex]
[tex]\[ k < -\frac{1}{4} \][/tex]
So, the system has no real solutions for:
[tex]\[ (-\infty, -\frac{1}{4}) \][/tex]
#### For one real solution: [tex]\(\Delta = 0\)[/tex]
[tex]\[ 1 + 4k = 0 \][/tex]
[tex]\[ 4k = -1 \][/tex]
[tex]\[ k = -\frac{1}{4} \][/tex]
So, the system has one real solution for:
[tex]\[ k = -\frac{1}{4} \][/tex]
#### For two real solutions: [tex]\(\Delta > 0\)[/tex]
[tex]\[ 1 + 4k > 0 \][/tex]
[tex]\[ 4k > -1 \][/tex]
[tex]\[ k > -\frac{1}{4} \][/tex]
So, the system has two real solutions for:
[tex]\[ (-\frac{1}{4}, \infty) \][/tex]
### Summary
- The system has no real solutions for [tex]\( k \)[/tex] in the interval [tex]\( (-\infty, -\frac{1}{4}) \)[/tex].
- The system has one real solution for [tex]\( k = -\frac{1}{4} \)[/tex].
- The system has two real solutions for [tex]\( k \)[/tex] in the interval [tex]\( (-\frac{1}{4}, \infty) \)[/tex].
In equations:
[tex]\[ \text{No real solutions: } k \in (-\infty, -\frac{1}{4}) \][/tex]
[tex]\[ \text{One real solution: } k = -\frac{1}{4} \][/tex]
[tex]\[ \text{Two real solutions: } k \in (-\frac{1}{4}, \infty) \][/tex]
### Given System of Equations:
[tex]\[ \begin{cases} y = x^2 \\ y = x + k \end{cases} \][/tex]
### Step 1: Set the equations equal to each other
Since both equations are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ x^2 = x + k \][/tex]
### Step 2: Rearrange into standard quadratic form
Rearrange the equation to the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ x^2 - x - k = 0 \][/tex]
### Step 3: Determine the discriminant
The discriminant of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For our equation [tex]\(x^2 - x - k = 0\)[/tex], we have [tex]\(a = 1\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -k\)[/tex]. Substituting these values into the formula for the discriminant, we get:
[tex]\[ \Delta = (-1)^2 - 4(1)(-k) = 1 + 4k \][/tex]
### Step 4: Analyze the discriminant
The number of real solutions to the quadratic equation depends on the value of the discriminant [tex]\(\Delta\)[/tex]:
- No real solutions: [tex]\(\Delta < 0\)[/tex]
- One real solution: [tex]\(\Delta = 0\)[/tex]
- Two real solutions: [tex]\(\Delta > 0\)[/tex]
#### For no real solutions: [tex]\(\Delta < 0\)[/tex]
[tex]\[ 1 + 4k < 0 \][/tex]
[tex]\[ 4k < -1 \][/tex]
[tex]\[ k < -\frac{1}{4} \][/tex]
So, the system has no real solutions for:
[tex]\[ (-\infty, -\frac{1}{4}) \][/tex]
#### For one real solution: [tex]\(\Delta = 0\)[/tex]
[tex]\[ 1 + 4k = 0 \][/tex]
[tex]\[ 4k = -1 \][/tex]
[tex]\[ k = -\frac{1}{4} \][/tex]
So, the system has one real solution for:
[tex]\[ k = -\frac{1}{4} \][/tex]
#### For two real solutions: [tex]\(\Delta > 0\)[/tex]
[tex]\[ 1 + 4k > 0 \][/tex]
[tex]\[ 4k > -1 \][/tex]
[tex]\[ k > -\frac{1}{4} \][/tex]
So, the system has two real solutions for:
[tex]\[ (-\frac{1}{4}, \infty) \][/tex]
### Summary
- The system has no real solutions for [tex]\( k \)[/tex] in the interval [tex]\( (-\infty, -\frac{1}{4}) \)[/tex].
- The system has one real solution for [tex]\( k = -\frac{1}{4} \)[/tex].
- The system has two real solutions for [tex]\( k \)[/tex] in the interval [tex]\( (-\frac{1}{4}, \infty) \)[/tex].
In equations:
[tex]\[ \text{No real solutions: } k \in (-\infty, -\frac{1}{4}) \][/tex]
[tex]\[ \text{One real solution: } k = -\frac{1}{4} \][/tex]
[tex]\[ \text{Two real solutions: } k \in (-\frac{1}{4}, \infty) \][/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
