Question
Determine the zero(s) of the quadratic function y=6x2+13x+5. Show all steps in easy form (2 points
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Answer
Put this in paragraph form
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Put it in easy words
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Make the number keyboard font
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Keyboard font put in paragraph form
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Easier words? And simple explanation?
Answered by
GPT-4o mini
To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). In this case:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
### Step 1: Compute the Discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = 13^2 = 169
\]
\[
4ac = 4 \times 6 \times 5 = 120
\]
Now, subtract \( 4ac \) from \( b^2 \):
\[
b^2 - 4ac = 169 - 120 = 49
\]
### Step 2: Apply the Quadratic Formula
Now that we have the discriminant, we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
x = \frac{{-13 \pm \sqrt{49}}}{{2 \cdot 6}}
\]
Calculating \( \sqrt{49} = 7 \):
\[
x = \frac{{-13 \pm 7}}{{12}}
\]
### Step 3: Solve for the Two Values of \( x \)
Now split into the two possible solutions:
1. \( x_1 = \frac{{-13 + 7}}{{12}} = \frac{{-6}}{{12}} = -\frac{1}{2} \)
2. \( x_2 = \frac{{-13 - 7}}{{12}} = \frac{{-20}}{{12}} = -\frac{5}{3} \)
### Summary of the Zeros
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). In this case:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
### Step 1: Compute the Discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = 13^2 = 169
\]
\[
4ac = 4 \times 6 \times 5 = 120
\]
Now, subtract \( 4ac \) from \( b^2 \):
\[
b^2 - 4ac = 169 - 120 = 49
\]
### Step 2: Apply the Quadratic Formula
Now that we have the discriminant, we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
x = \frac{{-13 \pm \sqrt{49}}}{{2 \cdot 6}}
\]
Calculating \( \sqrt{49} = 7 \):
\[
x = \frac{{-13 \pm 7}}{{12}}
\]
### Step 3: Solve for the Two Values of \( x \)
Now split into the two possible solutions:
1. \( x_1 = \frac{{-13 + 7}}{{12}} = \frac{{-6}}{{12}} = -\frac{1}{2} \)
2. \( x_2 = \frac{{-13 - 7}}{{12}} = \frac{{-20}}{{12}} = -\frac{5}{3} \)
### Summary of the Zeros
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]
Answered by
GPT-4o mini
To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we utilize the quadratic formula, which is given by \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \). In this case, the coefficients are \( a = 6 \), \( b = 13 \), and \( c = 5 \). First, we calculate the discriminant \( b^2 - 4ac \). Calculating \( b^2 \) gives us \( 13^2 = 169 \), and for \( 4ac \), we find \( 4 \times 6 \times 5 = 120 \). Thus, the discriminant becomes \( 169 - 120 = 49 \).
Next, we substitute \( a \), \( b \), and the discriminant into the quadratic formula. This gives us \( x = \frac{{-13 \pm \sqrt{49}}}{{12}} \). Since \( \sqrt{49} = 7 \), the expression simplifies to \( x = \frac{{-13 \pm 7}}{{12}} \). This results in two possible solutions for \( x \). The first solution, \( x_1 \), yields \( \frac{{-13 + 7}}{{12}} = -\frac{1}{2} \), while the second solution, \( x_2 \), results in \( \frac{{-13 - 7}}{{12}} = -\frac{20}{12} = -\frac{5}{3} \). Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).
Next, we substitute \( a \), \( b \), and the discriminant into the quadratic formula. This gives us \( x = \frac{{-13 \pm \sqrt{49}}}{{12}} \). Since \( \sqrt{49} = 7 \), the expression simplifies to \( x = \frac{{-13 \pm 7}}{{12}} \). This results in two possible solutions for \( x \). The first solution, \( x_1 \), yields \( \frac{{-13 + 7}}{{12}} = -\frac{1}{2} \), while the second solution, \( x_2 \), results in \( \frac{{-13 - 7}}{{12}} = -\frac{20}{12} = -\frac{5}{3} \). Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).
Answered by
GPT-4o mini
To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula, which is written as \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \). In this equation, \( a \), \( b \), and \( c \) are the numbers from our quadratic equation, where \( a = 6 \), \( b = 13 \), and \( c = 5 \).
First, we need to calculate something called the discriminant, which is found using the formula \( b^2 - 4ac \). For our equation, we calculate \( b^2 \), which is \( 13^2 = 169 \), and \( 4ac \), which equals \( 4 \times 6 \times 5 = 120 \). Now, we subtract \( 120 \) from \( 169 \) to get the discriminant: \( 169 - 120 = 49 \).
Now we can plug our values into the quadratic formula. This gives us \( x = \frac{{-13 \pm \sqrt{49}}}{{12}} \). Since the square root of \( 49 \) is \( 7 \), we can rewrite the formula as \( x = \frac{{-13 \pm 7}}{{12}} \). This leads us to two possible values for \( x \):
1. For the first solution, we calculate \( x_1 = \frac{{-13 + 7}}{{12}} = -\frac{1}{2} \).
2. For the second solution, we find \( x_2 = \frac{{-13 - 7}}{{12}} = -\frac{20}{12} = -\frac{5}{3} \).
Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).
First, we need to calculate something called the discriminant, which is found using the formula \( b^2 - 4ac \). For our equation, we calculate \( b^2 \), which is \( 13^2 = 169 \), and \( 4ac \), which equals \( 4 \times 6 \times 5 = 120 \). Now, we subtract \( 120 \) from \( 169 \) to get the discriminant: \( 169 - 120 = 49 \).
Now we can plug our values into the quadratic formula. This gives us \( x = \frac{{-13 \pm \sqrt{49}}}{{12}} \). Since the square root of \( 49 \) is \( 7 \), we can rewrite the formula as \( x = \frac{{-13 \pm 7}}{{12}} \). This leads us to two possible values for \( x \):
1. For the first solution, we calculate \( x_1 = \frac{{-13 + 7}}{{12}} = -\frac{1}{2} \).
2. For the second solution, we find \( x_2 = \frac{{-13 - 7}}{{12}} = -\frac{20}{12} = -\frac{5}{3} \).
Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).
Answered by
GPT-4o mini
Sure! Here’s the explanation using a number keyboard font style for the equations:
To find the zeros of the quadratic function **y = 6x² + 13x + 5**, we can use the quadratic formula, which is written as
```
x = (-b ± √(b² - 4ac)) / (2a)
```
In this equation, **a**, **b**, and **c** are the numbers from our quadratic equation, where **a = 6**, **b = 13**, and **c = 5**.
First, we need to calculate something called the *discriminant*, which is found using the formula
```
b² - 4ac.
```
For our equation, we calculate **b²**, which is
```
13² = 169,
```
and **4ac**, which equals
```
4 × 6 × 5 = 120.
```
Now, we subtract **120** from **169** to get the discriminant:
```
169 - 120 = 49.
```
Now we can plug our values into the quadratic formula. This gives us
```
x = (-13 ± √49) / 12.
```
Since the square root of **49** is **7**, we can rewrite the formula as
```
x = (-13 ± 7) / 12.
```
This leads us to two possible values for **x**:
1. For the first solution, we calculate
```
x₁ = (-13 + 7) / 12 = -1/2.
```
2. For the second solution, we find
```
x₂ = (-13 - 7) / 12 = -20/12 = -5/3.
```
Therefore, the zeros of the quadratic function **y = 6x² + 13x + 5** are
```
x = -1/2
```
and
```
x = -5/3.
```
Feel free to let me know if there's anything else you need!
To find the zeros of the quadratic function **y = 6x² + 13x + 5**, we can use the quadratic formula, which is written as
```
x = (-b ± √(b² - 4ac)) / (2a)
```
In this equation, **a**, **b**, and **c** are the numbers from our quadratic equation, where **a = 6**, **b = 13**, and **c = 5**.
First, we need to calculate something called the *discriminant*, which is found using the formula
```
b² - 4ac.
```
For our equation, we calculate **b²**, which is
```
13² = 169,
```
and **4ac**, which equals
```
4 × 6 × 5 = 120.
```
Now, we subtract **120** from **169** to get the discriminant:
```
169 - 120 = 49.
```
Now we can plug our values into the quadratic formula. This gives us
```
x = (-13 ± √49) / 12.
```
Since the square root of **49** is **7**, we can rewrite the formula as
```
x = (-13 ± 7) / 12.
```
This leads us to two possible values for **x**:
1. For the first solution, we calculate
```
x₁ = (-13 + 7) / 12 = -1/2.
```
2. For the second solution, we find
```
x₂ = (-13 - 7) / 12 = -20/12 = -5/3.
```
Therefore, the zeros of the quadratic function **y = 6x² + 13x + 5** are
```
x = -1/2
```
and
```
x = -5/3.
```
Feel free to let me know if there's anything else you need!
Answered by
GPT-4o mini
Sure! Here’s the explanation in paragraph form using keyboard-style formatting for the equations:
To find the zeros of the quadratic function **y = 6x² + 13x + 5**, we can use the quadratic formula, which is written as **x = (-b ± √(b² - 4ac)) / (2a)**. In this equation, **a**, **b**, and **c** are the numbers from our quadratic equation, where **a = 6**, **b = 13**, and **c = 5**. First, we need to calculate something called the *discriminant*, which is found using the formula **b² - 4ac**. For our equation, we calculate **b²**, which is **13² = 169**, and **4ac**, which equals **4 × 6 × 5 = 120**. Now, we subtract **120** from **169** to get the discriminant: **169 - 120 = 49**.
Next, we can plug our values into the quadratic formula. This gives us **x = (-13 ± √49) / 12**. Since the square root of **49** is **7**, we can rewrite the formula as **x = (-13 ± 7) / 12**. This leads us to two possible values for **x**: for the first solution, we calculate **x₁ = (-13 + 7) / 12 = -1/2**. For the second solution, we find **x₂ = (-13 - 7) / 12 = -20/12 = -5/3**. Therefore, the zeros of the quadratic function **y = 6x² + 13x + 5** are **x = -1/2** and **x = -5/3**.
To find the zeros of the quadratic function **y = 6x² + 13x + 5**, we can use the quadratic formula, which is written as **x = (-b ± √(b² - 4ac)) / (2a)**. In this equation, **a**, **b**, and **c** are the numbers from our quadratic equation, where **a = 6**, **b = 13**, and **c = 5**. First, we need to calculate something called the *discriminant*, which is found using the formula **b² - 4ac**. For our equation, we calculate **b²**, which is **13² = 169**, and **4ac**, which equals **4 × 6 × 5 = 120**. Now, we subtract **120** from **169** to get the discriminant: **169 - 120 = 49**.
Next, we can plug our values into the quadratic formula. This gives us **x = (-13 ± √49) / 12**. Since the square root of **49** is **7**, we can rewrite the formula as **x = (-13 ± 7) / 12**. This leads us to two possible values for **x**: for the first solution, we calculate **x₁ = (-13 + 7) / 12 = -1/2**. For the second solution, we find **x₂ = (-13 - 7) / 12 = -20/12 = -5/3**. Therefore, the zeros of the quadratic function **y = 6x² + 13x + 5** are **x = -1/2** and **x = -5/3**.
Answered by
GPT-4o mini
Sure! Here’s a simpler explanation in easier words:
To find the zeros of the function \( y = 6x^2 + 13x + 5 \), we use a formula called the quadratic formula, which looks like this:
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
Here, **a**, **b**, and **c** are the numbers from our equation, where **a = 6**, **b = 13**, and **c = 5**.
First, we need to find something called the discriminant. This helps us understand if there are any real solutions. We calculate the discriminant using the formula \( b^2 - 4ac \). For our numbers, we find:
- \( b^2 = 13^2 = 169 \)
- \( 4ac = 4 \times 6 \times 5 = 120 \)
Now we subtract to find the discriminant:
\[
169 - 120 = 49
\]
Next, we plug our numbers into the quadratic formula. This gives us:
\[
x = \frac{{-13 \pm \sqrt{49}}}{12}
\]
Since the square root of **49** is **7**, we rewrite it as:
\[
x = \frac{{-13 \pm 7}}{12}
\]
Now we can find our two solutions for \( x \):
1. For the first solution, we do:
\[
x_1 = \frac{{-13 + 7}}{12} = \frac{{-6}}{12} = -\frac{1}{2}
\]
2. For the second solution, we do:
\[
x_2 = \frac{{-13 - 7}}{12} = \frac{{-20}}{12} = -\frac{5}{3}
\]
So, the zeros (or the solutions) of the function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \). This means these are the points where the graph of the function touches the x-axis.
To find the zeros of the function \( y = 6x^2 + 13x + 5 \), we use a formula called the quadratic formula, which looks like this:
\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
\]
Here, **a**, **b**, and **c** are the numbers from our equation, where **a = 6**, **b = 13**, and **c = 5**.
First, we need to find something called the discriminant. This helps us understand if there are any real solutions. We calculate the discriminant using the formula \( b^2 - 4ac \). For our numbers, we find:
- \( b^2 = 13^2 = 169 \)
- \( 4ac = 4 \times 6 \times 5 = 120 \)
Now we subtract to find the discriminant:
\[
169 - 120 = 49
\]
Next, we plug our numbers into the quadratic formula. This gives us:
\[
x = \frac{{-13 \pm \sqrt{49}}}{12}
\]
Since the square root of **49** is **7**, we rewrite it as:
\[
x = \frac{{-13 \pm 7}}{12}
\]
Now we can find our two solutions for \( x \):
1. For the first solution, we do:
\[
x_1 = \frac{{-13 + 7}}{12} = \frac{{-6}}{12} = -\frac{1}{2}
\]
2. For the second solution, we do:
\[
x_2 = \frac{{-13 - 7}}{12} = \frac{{-20}}{12} = -\frac{5}{3}
\]
So, the zeros (or the solutions) of the function \( y = 6x^2 + 13x + 5 \) are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \). This means these are the points where the graph of the function touches the x-axis.