To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we will follow the steps of using the quadratic formula.
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Identify the coefficients: The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). For our function, we can identify:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
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Write the quadratic formula: The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
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Substitute the values into the formula: We will substitute the values of \( a \), \( b \), and \( c \) into the formula: \[ x = \frac{-13 \pm \sqrt{13^2 - 4 \times 6 \times 5}}{2 \times 6} \]
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Calculate \( b^2 - 4ac \): First, we need to calculate \( 13^2 \):
- \( 13^2 = 169 \)
Next, calculate \( 4ac \):
- \( 4 \times 6 = 24 \)
- Now, multiply by \( c \):
- \( 24 \times 5 = 120 \)
Now we subtract: \[ b^2 - 4ac = 169 - 120 = 49 \]
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Continue substituting back into the formula: Now we substitute back into the quadratic formula: \[ x = \frac{-13 \pm \sqrt{49}}{12} \]
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Calculate the square root: The square root of \( 49 \) is \( 7 \): \[ x = \frac{-13 \pm 7}{12} \]
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Find the two potential solutions: We will calculate the two values from the plus and minus:
- Using the plus sign: \[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Using the minus sign: \[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
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State the zeros of the quadratic function: Therefore, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
- \( x = -\frac{1}{2} \)
- \( x = -\frac{5}{3} \)
In conclusion, the zeros of the function are \( -\frac{1}{2} \) and \( -\frac{5}{3} \).