Answer :
Certainly! Let's break down the problem step-by-step.
### Step 1: Define the Quadratic Equation
We are given the quadratic equation:
[tex]\[ y = x^2 + 4x - 5 \][/tex]
### Step 2: Finding the Roots of the Equation
To find the roots of the equation (the values of [tex]\(x\)[/tex] for which [tex]\(y = 0\)[/tex]), we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( y = x^2 + 4x - 5 \)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 4\)[/tex]
- [tex]\(c = -5\)[/tex]
#### Calculate the Discriminant
First, we need to calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = 4^2 - 4 \cdot 1 \cdot (-5) \][/tex]
[tex]\[ \Delta = 16 + 20 \][/tex]
[tex]\[ \Delta = 36 \][/tex]
#### Find the Roots
Using the discriminant, we can now find the roots:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values:
[tex]\[ x_1 = \frac{-4 + \sqrt{36}}{2 \cdot 1} \][/tex]
[tex]\[ x_1 = \frac{-4 + 6}{2} \][/tex]
[tex]\[ x_1 = \frac{2}{2} \][/tex]
[tex]\[ x_1 = 1 \][/tex]
[tex]\[ x_2 = \frac{-4 - \sqrt{36}}{2 \cdot 1} \][/tex]
[tex]\[ x_2 = \frac{-4 - 6}{2} \][/tex]
[tex]\[ x_2 = \frac{-10}{2} \][/tex]
[tex]\[ x_2 = -5 \][/tex]
So, the roots of the equation are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
### Step 3: Evaluate [tex]\( y \)[/tex] at Various [tex]\( x \)[/tex] Values
Let's evaluate [tex]\( y \)[/tex] at a few chosen values of [tex]\( x \)[/tex]. We will compute [tex]\( y \)[/tex] for [tex]\( x = -2, 0, 1, 3 \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = (-2)^2 + 4(-2) - 5 \][/tex]
[tex]\[ y = 4 - 8 - 5 \][/tex]
[tex]\[ y = -9 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = (0)^2 + 4(0) - 5 \][/tex]
[tex]\[ y = -5 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
Since we already know one of the roots is [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1^2 + 4 \cdot 1 - 5 \][/tex]
[tex]\[ y = 1 + 4 - 5 \][/tex]
[tex]\[ y = 0 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3^2 + 4 \cdot 3 - 5 \][/tex]
[tex]\[ y = 9 + 12 - 5 \][/tex]
[tex]\[ y = 16 \][/tex]
### Final Results
Summarizing the results:
- The discriminant is [tex]\( 36 \)[/tex].
- The roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
- The values of [tex]\( y \)[/tex] at the evaluated [tex]\( x \)[/tex] values are:
- At [tex]\( x = -2 \)[/tex], [tex]\( y = -9 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = -5 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 3 \)[/tex], [tex]\( y = 16 \)[/tex]
Thus, these steps provide a detailed solution to the given quadratic equation problem.
### Step 1: Define the Quadratic Equation
We are given the quadratic equation:
[tex]\[ y = x^2 + 4x - 5 \][/tex]
### Step 2: Finding the Roots of the Equation
To find the roots of the equation (the values of [tex]\(x\)[/tex] for which [tex]\(y = 0\)[/tex]), we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\( y = x^2 + 4x - 5 \)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 4\)[/tex]
- [tex]\(c = -5\)[/tex]
#### Calculate the Discriminant
First, we need to calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = 4^2 - 4 \cdot 1 \cdot (-5) \][/tex]
[tex]\[ \Delta = 16 + 20 \][/tex]
[tex]\[ \Delta = 36 \][/tex]
#### Find the Roots
Using the discriminant, we can now find the roots:
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
Substituting the values:
[tex]\[ x_1 = \frac{-4 + \sqrt{36}}{2 \cdot 1} \][/tex]
[tex]\[ x_1 = \frac{-4 + 6}{2} \][/tex]
[tex]\[ x_1 = \frac{2}{2} \][/tex]
[tex]\[ x_1 = 1 \][/tex]
[tex]\[ x_2 = \frac{-4 - \sqrt{36}}{2 \cdot 1} \][/tex]
[tex]\[ x_2 = \frac{-4 - 6}{2} \][/tex]
[tex]\[ x_2 = \frac{-10}{2} \][/tex]
[tex]\[ x_2 = -5 \][/tex]
So, the roots of the equation are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
### Step 3: Evaluate [tex]\( y \)[/tex] at Various [tex]\( x \)[/tex] Values
Let's evaluate [tex]\( y \)[/tex] at a few chosen values of [tex]\( x \)[/tex]. We will compute [tex]\( y \)[/tex] for [tex]\( x = -2, 0, 1, 3 \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = (-2)^2 + 4(-2) - 5 \][/tex]
[tex]\[ y = 4 - 8 - 5 \][/tex]
[tex]\[ y = -9 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = (0)^2 + 4(0) - 5 \][/tex]
[tex]\[ y = -5 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
Since we already know one of the roots is [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1^2 + 4 \cdot 1 - 5 \][/tex]
[tex]\[ y = 1 + 4 - 5 \][/tex]
[tex]\[ y = 0 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3^2 + 4 \cdot 3 - 5 \][/tex]
[tex]\[ y = 9 + 12 - 5 \][/tex]
[tex]\[ y = 16 \][/tex]
### Final Results
Summarizing the results:
- The discriminant is [tex]\( 36 \)[/tex].
- The roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = -5 \)[/tex].
- The values of [tex]\( y \)[/tex] at the evaluated [tex]\( x \)[/tex] values are:
- At [tex]\( x = -2 \)[/tex], [tex]\( y = -9 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = -5 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 3 \)[/tex], [tex]\( y = 16 \)[/tex]
Thus, these steps provide a detailed solution to the given quadratic equation problem.
