To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can follow these steps:
Step One: Identify the coefficients
In the quadratic expression \( ax^2 + bx + c \):
- The coefficient \( a \) is six.
- The coefficient \( b \) is thirteen.
- The coefficient \( c \) is five.
Step Two: Use the quadratic formula
The formula to find the zeros of a quadratic function is given by: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] We will substitute the values of \( a \), \( b \), and \( c \) into this formula.
Step Three: Calculate the discriminant
We need to calculate the discriminant, which is \( b^2 - 4ac \):
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Calculate \( b^2 \): \[ b^2 = 13^2 = 169 \]
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Calculate \( 4ac \): \[ 4ac = 4 \times 6 \times 5 = 120 \]
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Now, subtract \( 4ac \) from \( b^2 \): \[ b^2 - 4ac = 169 - 120 = 49 \]
Step Four: Substitute back into the quadratic formula
Now that we have \( b^2 - 4ac = 49 \), we can find \( \sqrt{49} \): \[ \sqrt{49} = 7 \]
Substituting these values into the quadratic formula gives: \[ x = \frac{{-13 \pm 7}}{2 \times 6} \] This simplifies to: \[ x = \frac{{-13 \pm 7}}{12} \]
Step Five: Calculate both solutions
Now we will calculate the two possible values for \( x \):
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First value: \[ x = \frac{{-13 + 7}}{12} = \frac{{-6}}{12} = -\frac{1}{2} \]
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Second value: \[ x = \frac{{-13 - 7}}{12} = \frac{{-20}}{12} = -\frac{5}{3} \]
Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
- \( x = -\frac{1}{2} \)
- \( x = -\frac{5}{3} \)
Thus, the solutions to the equation are \( x = -0.5 \) and \( x \approx -1.67 \).