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Determine the zero(s) of the quadratic function y=6x2+13x+5. Show all steps.Question
Determine the zero(s) of the quadratic function y=6x2+13x+5
. Show all steps
. Show all steps
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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 \( a = 6 \), \( b = 13 \), and \( c = 5 \).
### Step 1: Identify coefficients
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
### Step 2: Calculate the discriminant
The discriminant (\( D \)) is given by:
\[
D = b^2 - 4ac
\]
Substituting in the values of \( a \), \( b \), and \( c \):
\[
D = 13^2 - 4 \cdot 6 \cdot 5
\]
\[
D = 169 - 120
\]
\[
D = 49
\]
### Step 3: Find the roots using the quadratic formula
Now that we have calculated the discriminant, we can substitute it back into the quadratic formula:
\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
Calculating \( \sqrt{49} \):
\[
\sqrt{49} = 7
\]
Now substituting this value back into the equation:
\[
x = \frac{-13 \pm 7}{12}
\]
### Step 4: Calculate the two possible values for \( x \)
**First root:**
\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
**Second root:**
\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]
### Step 5: State the zeros of the function
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 \( a = 6 \), \( b = 13 \), and \( c = 5 \).
### Step 1: Identify coefficients
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
### Step 2: Calculate the discriminant
The discriminant (\( D \)) is given by:
\[
D = b^2 - 4ac
\]
Substituting in the values of \( a \), \( b \), and \( c \):
\[
D = 13^2 - 4 \cdot 6 \cdot 5
\]
\[
D = 169 - 120
\]
\[
D = 49
\]
### Step 3: Find the roots using the quadratic formula
Now that we have calculated the discriminant, we can substitute it back into the quadratic formula:
\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
Calculating \( \sqrt{49} \):
\[
\sqrt{49} = 7
\]
Now substituting this value back into the equation:
\[
x = \frac{-13 \pm 7}{12}
\]
### Step 4: Calculate the two possible values for \( x \)
**First root:**
\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
**Second root:**
\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]
### Step 5: State the zeros of the function
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]
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