Answer :
Sure, let's solve the quadratic equation [tex]\(3x^2 + x - 5 = 0\)[/tex] step-by-step.
### Step 1: Identify the coefficients
In the quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the coefficients are:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -5\)[/tex]
### Step 2: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 1^2 - 4 \cdot 3 \cdot (-5) = 1 + 60 = 61 \][/tex]
### Step 3: Use the quadratic formula
The quadratic formula to find the roots of [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{61}}{2 \cdot 3} \][/tex]
### Step 4: Calculate the two solutions
Now, we'll find the two solutions by considering the [tex]\(+\)[/tex] and [tex]\(-\)[/tex] in the quadratic formula separately:
#### Solution 1:
[tex]\[ x_1 = \frac{-1 + \sqrt{61}}{6} \][/tex]
After performing the calculations, we get:
[tex]\[ x_1 \approx 1.135041612651109 \][/tex]
#### Solution 2:
[tex]\[ x_2 = \frac{-1 - \sqrt{61}}{6} \][/tex]
After performing the calculations, we get:
[tex]\[ x_2 \approx -1.4683749459844424 \][/tex]
### Step 5: Round the solutions to two decimal places
Rounding the solutions, we have:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
### Final Answer
The solutions to the quadratic equation [tex]\(3x^2 + x - 5 = 0\)[/tex], rounded to two decimal places, are:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
### Step 1: Identify the coefficients
In the quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the coefficients are:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -5\)[/tex]
### Step 2: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 1^2 - 4 \cdot 3 \cdot (-5) = 1 + 60 = 61 \][/tex]
### Step 3: Use the quadratic formula
The quadratic formula to find the roots of [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{61}}{2 \cdot 3} \][/tex]
### Step 4: Calculate the two solutions
Now, we'll find the two solutions by considering the [tex]\(+\)[/tex] and [tex]\(-\)[/tex] in the quadratic formula separately:
#### Solution 1:
[tex]\[ x_1 = \frac{-1 + \sqrt{61}}{6} \][/tex]
After performing the calculations, we get:
[tex]\[ x_1 \approx 1.135041612651109 \][/tex]
#### Solution 2:
[tex]\[ x_2 = \frac{-1 - \sqrt{61}}{6} \][/tex]
After performing the calculations, we get:
[tex]\[ x_2 \approx -1.4683749459844424 \][/tex]
### Step 5: Round the solutions to two decimal places
Rounding the solutions, we have:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
### Final Answer
The solutions to the quadratic equation [tex]\(3x^2 + x - 5 = 0\)[/tex], rounded to two decimal places, are:
[tex]\[ x_1 \approx 1.14 \][/tex]
[tex]\[ x_2 \approx -1.47 \][/tex]
