Answer :
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]
