Question
find the zeroes y-6x^2+13x+5 show your work
Answers
GPT-4o mini
To find the zeroes of the quadratic equation \( y - 6x^2 + 13x + 5 = 0 \), we first set the equation to zero:
\[
-6x^2 + 13x + 5 = 0.
\]
To make it easier to work with, we could multiply the entire equation by -1:
\[
6x^2 - 13x - 5 = 0.
\]
Now, we will use the quadratic formula to find the roots of the equation, where the quadratic equation is given by \( ax^2 + bx + c = 0 \). In our case, \( a = 6 \), \( b = -13 \), and \( c = -5 \).
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]
1. **Calculate the discriminant** (\( b^2 - 4ac \)):
\[
b^2 = (-13)^2 = 169,
\]
\[
4ac = 4 \cdot 6 \cdot (-5) = -120.
\]
So,
\[
b^2 - 4ac = 169 - (-120) = 169 + 120 = 289.
\]
2. **Plug values into the formula**:
Now substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
x = \frac{-(-13) \pm \sqrt{289}}{2 \cdot 6}.
\]
Calculating the square root:
\[
\sqrt{289} = 17.
\]
Substituting this back into the formula:
\[
x = \frac{13 \pm 17}{12}.
\]
3. **Solve for the two possible values of \( x \)**:
**First root:**
\[
x = \frac{13 + 17}{12} = \frac{30}{12} = \frac{5}{2} = 2.5.
\]
**Second root:**
\[
x = \frac{13 - 17}{12} = \frac{-4}{12} = \frac{-1}{3}.
\]
So the zeroes of the equation \( y - 6x^2 + 13x + 5 = 0 \) are:
\[
x = 2.5 \quad \text{and} \quad x = -\frac{1}{3}.
\]
\[
-6x^2 + 13x + 5 = 0.
\]
To make it easier to work with, we could multiply the entire equation by -1:
\[
6x^2 - 13x - 5 = 0.
\]
Now, we will use the quadratic formula to find the roots of the equation, where the quadratic equation is given by \( ax^2 + bx + c = 0 \). In our case, \( a = 6 \), \( b = -13 \), and \( c = -5 \).
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]
1. **Calculate the discriminant** (\( b^2 - 4ac \)):
\[
b^2 = (-13)^2 = 169,
\]
\[
4ac = 4 \cdot 6 \cdot (-5) = -120.
\]
So,
\[
b^2 - 4ac = 169 - (-120) = 169 + 120 = 289.
\]
2. **Plug values into the formula**:
Now substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[
x = \frac{-(-13) \pm \sqrt{289}}{2 \cdot 6}.
\]
Calculating the square root:
\[
\sqrt{289} = 17.
\]
Substituting this back into the formula:
\[
x = \frac{13 \pm 17}{12}.
\]
3. **Solve for the two possible values of \( x \)**:
**First root:**
\[
x = \frac{13 + 17}{12} = \frac{30}{12} = \frac{5}{2} = 2.5.
\]
**Second root:**
\[
x = \frac{13 - 17}{12} = \frac{-4}{12} = \frac{-1}{3}.
\]
So the zeroes of the equation \( y - 6x^2 + 13x + 5 = 0 \) are:
\[
x = 2.5 \quad \text{and} \quad x = -\frac{1}{3}.
\]