To find the zeros of the quadratic function \( y = x^2 + 16x - 39 \), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 16 \), and \( c = -39 \).
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 16^2 = 256 \] \[ 4ac = 4 \cdot 1 \cdot (-39) = -156 \] \[ b^2 - 4ac = 256 - (-156) = 256 + 156 = 412 \]
- Calculate the roots using the quadratic formula:
\[ x = \frac{-16 \pm \sqrt{412}}{2 \cdot 1} \]
- Simplify \( \sqrt{412} \):
First, factor \( 412 \):
\[ 412 = 4 \cdot 103 \]
So,
\[ \sqrt{412} = \sqrt{4 \cdot 103} = 2\sqrt{103} \]
- Substituting back into the quadratic formula:
\[ x = \frac{-16 \pm 2\sqrt{103}}{2} \] \[ x = -8 \pm \sqrt{103} \]
- Finding the two zeros:
- Smaller zero:
\[ x = -8 - \sqrt{103} \]
- Greater zero:
\[ x = -8 + \sqrt{103} \]
The exact values of the zeros are:
- Smaller zero: \( x = -8 - \sqrt{103} \)
- Greater zero: \( x = -8 + \sqrt{103} \)
Now, we can approximate \( \sqrt{103} \):
\[ \sqrt{103} \approx 10.14 \]
Then we find:
- Smaller zero \(\approx -8 - 10.14 \approx -18.14\)
- Greater zero \(\approx -8 + 10.14 \approx 2.14\)
Thus, the answers are:
- Smaller zero: \( x \approx -18.14 \) (exactly \( x = -8 - \sqrt{103} \))
- Greater zero: \( x \approx 2.14 \) (exactly \( x = -8 + \sqrt{103} \))
To summarize:
- smaller zero: \( x = -8 - \sqrt{103} \)
- greater zero: \( x = -8 + \sqrt{103} \)
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