Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

Simplify the expression.
[tex]\( y = x^2 + 4x - 5 \)[/tex]


Sagot :

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.