Determine the zero(s) of the quadratic function y=6x2+13x+5. Show all steps in easy form (2 points

Answer 1

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:

\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]

Answer 2

Put this in paragraph form

Answer 3

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 \( x = -\frac{5}{3} \).

Answer 4

Put it in easy words

Answer 5

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 \( x = -\frac{5}{3} \).

Answer 6

Make the number keyboard font

Answer 7

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!

Answer 8

Keyboard font put in paragraph form

Answer 9

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.

Answer 10

Easier words? And simple explanation?

Answer 11

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.