Question
Find the zeros of the quadratic equation y=6x2+13x+5 . You may use any of the following methods: complete the square
Answers
Answered by
GPT-4o mini
To find the zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) by completing the square, follow these steps:
1. **Rewrite the equation** in the standard form (if necessary):
\[
y = 6x^2 + 13x + 5
\]
2. **Factor out the coefficient of \( x^2 \)** from the first two terms:
\[
y = 6\left(x^2 + \frac{13}{6}x\right) + 5
\]
3. **Complete the square** inside the parentheses. To do this, take half of the coefficient of \( x \) (which is \( \frac{13}{6} \)), square it, and add and subtract this value inside the parentheses:
- Half of \( \frac{13}{6} \) is \( \frac{13}{12} \).
- Squaring \( \frac{13}{12} \) gives:
\[
\left(\frac{13}{12}\right)^2 = \frac{169}{144}
\]
4. Now, add and subtract this squared term inside the parentheses:
\[
y = 6\left(x^2 + \frac{13}{6}x + \frac{169}{144} - \frac{169}{144}\right) + 5
\]
5. Simplify the equation:
\[
y = 6\left(\left(x + \frac{13}{12}\right)^2 - \frac{169}{144}\right) + 5
\]
6. Distributing the \( 6 \):
\[
y = 6\left(x + \frac{13}{12}\right)^2 - 6 \cdot \frac{169}{144} + 5
\]
Calculate \( - 6 \cdot \frac{169}{144} \):
\[
- \frac{1014}{144} = -\frac{169}{24}
\]
Now, add \( 5 \) (which is \( \frac{120}{24} \)):
\[
-\frac{169}{24} + \frac{120}{24} = -\frac{49}{24}
\]
7. Now the equation is:
\[
y = 6\left(x + \frac{13}{12}\right)^2 - \frac{49}{24}
\]
8. **Set \( y = 0 \)** to find the zeros:
\[
6\left(x + \frac{13}{12}\right)^2 - \frac{49}{24} = 0
\]
Rearranging gives:
\[
6\left(x + \frac{13}{12}\right)^2 = \frac{49}{24}
\]
Dividing both sides by \( 6 \):
\[
\left(x + \frac{13}{12}\right)^2 = \frac{49}{144}
\]
9. **Take the square root** of both sides:
\[
x + \frac{13}{12} = \pm \frac{7}{12}
\]
10. **Solve for \( x \)**:
- For the positive root:
\[
x + \frac{13}{12} = \frac{7}{12} \implies x = \frac{7}{12} - \frac{13}{12} = -\frac{6}{12} = -\frac{1}{2}
\]
- For the negative root:
\[
x + \frac{13}{12} = -\frac{7}{12} \implies x = -\frac{7}{12} - \frac{13}{12} = -\frac{20}{12} = -\frac{5}{3}
\]
The zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) are:
\[
\boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{-\frac{5}{3}}
\]
1. **Rewrite the equation** in the standard form (if necessary):
\[
y = 6x^2 + 13x + 5
\]
2. **Factor out the coefficient of \( x^2 \)** from the first two terms:
\[
y = 6\left(x^2 + \frac{13}{6}x\right) + 5
\]
3. **Complete the square** inside the parentheses. To do this, take half of the coefficient of \( x \) (which is \( \frac{13}{6} \)), square it, and add and subtract this value inside the parentheses:
- Half of \( \frac{13}{6} \) is \( \frac{13}{12} \).
- Squaring \( \frac{13}{12} \) gives:
\[
\left(\frac{13}{12}\right)^2 = \frac{169}{144}
\]
4. Now, add and subtract this squared term inside the parentheses:
\[
y = 6\left(x^2 + \frac{13}{6}x + \frac{169}{144} - \frac{169}{144}\right) + 5
\]
5. Simplify the equation:
\[
y = 6\left(\left(x + \frac{13}{12}\right)^2 - \frac{169}{144}\right) + 5
\]
6. Distributing the \( 6 \):
\[
y = 6\left(x + \frac{13}{12}\right)^2 - 6 \cdot \frac{169}{144} + 5
\]
Calculate \( - 6 \cdot \frac{169}{144} \):
\[
- \frac{1014}{144} = -\frac{169}{24}
\]
Now, add \( 5 \) (which is \( \frac{120}{24} \)):
\[
-\frac{169}{24} + \frac{120}{24} = -\frac{49}{24}
\]
7. Now the equation is:
\[
y = 6\left(x + \frac{13}{12}\right)^2 - \frac{49}{24}
\]
8. **Set \( y = 0 \)** to find the zeros:
\[
6\left(x + \frac{13}{12}\right)^2 - \frac{49}{24} = 0
\]
Rearranging gives:
\[
6\left(x + \frac{13}{12}\right)^2 = \frac{49}{24}
\]
Dividing both sides by \( 6 \):
\[
\left(x + \frac{13}{12}\right)^2 = \frac{49}{144}
\]
9. **Take the square root** of both sides:
\[
x + \frac{13}{12} = \pm \frac{7}{12}
\]
10. **Solve for \( x \)**:
- For the positive root:
\[
x + \frac{13}{12} = \frac{7}{12} \implies x = \frac{7}{12} - \frac{13}{12} = -\frac{6}{12} = -\frac{1}{2}
\]
- For the negative root:
\[
x + \frac{13}{12} = -\frac{7}{12} \implies x = -\frac{7}{12} - \frac{13}{12} = -\frac{20}{12} = -\frac{5}{3}
\]
The zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) are:
\[
\boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{-\frac{5}{3}}
\]